Fill in the Blanks

• If a body changes its position with respect to its surroundings, the body is said to be in …………..

If a body changes its position with respect to its surroundings, the body is said to be in motion.

Motion is a relative concept.

• The distance can never be ………., while displacement can be.

The distance can never be negative, while displacement can be.

Distance is always positive while displacement can be positive or negative or zero.

• If a body moves such that it covers equal distances in equal intervals of time, whatsoever be small, then the object is said to have …………. motion.

If a body moves such that it covers equal distances in equal intervals of time, whatsoever be small, then the object is said to have uniform motion.

• When a particle moves in a straight line from point A to point B the distance covered is ………. the magnitude of the displacement.

When a particle moves in a straight line from point A to point B the distance covered is equal to the magnitude of the displacement.

• Distance travelled divided by elapsed time gives ………….

Distance travelled divided by elapsed time gives speed.

• The ratio of the total displacement of a body to the total time taken is …………

The ratio of the total displacement of a body to the total time taken is velocity.

The average velocity of an object moving on a circle of radius R in one complete rotation if it takes t second in completing one rotation is …………..

The average velocity of an object moving on a circle of radius R in one complete rotation if it takes t second in completing one rotation is zero.

In one complete rotation displacement of the object is zero

Velocity = Displacement/time = 0/t = 0

• An object moving uniformly on a circular track. Its velocity ………. with time.

An object moving uniformly on a circular track. Its velocity changes with time.

In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of velocity changes continuously. Hence body in circular motion moves with acceleration.

• If particle is moving along circular path such that in equal interval of time it describes the equal angle, then velocity vector …………

If particle is moving along circular path such that in equal interval of time it describes the equal angle, then velocity vector changes its direction continuously.

The particle describes the equal angle in equal interval of time. Hence particle is in uniform circular motion. In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of velocity changes continuously.

• The magnitude of average velocity ………… equal to the average speed.

The magnitude of average velocity may or may not be equal to the average speed.

• When a body has unequal displacement in equal intervals of time, it is said to be moving with …………..

When a body has unequal displacement in equal intervals of time, it is said to be moving with non-uniform velocity.

• The acceleration of freely falling object is approximately equal to ……..

The acceleration of freely falling object is approximately equal to 9.8 ms^-2.

• If a body starts from rest and moves with uniform acceleration, then its displacement is proportional to ………….

If a body starts from rest and moves with uniform acceleration, then its displacement is proportional to square of time.

s = ut + 1/2 at²

now u = 0

s = 1/2 at²

• A particle is just released to fall down from the top of a building. Its velocity at a particular position is directly proportional to …….

A particle is just released to fall down from the top of a building. Its velocity at a particular position is directly proportional to ………

A particle is just released to fall down from the top of a building. Its velocity at a particular position is directly proportional to time.

v = u + at

now u = 0

v = at

• All objects in free fall at a given place have the same

All objects in free fall at a given place have the same acceleration.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

• …………… acceleration is called retardation or deceleration.

Negative acceleration is called retardation or deceleration.

• A car is moving northward. The driver of the car applies brakes, the direction of acceleration is …………..

A car is moving northward. The driver of the car applies brakes, the direction of acceleration is southward.

When brakes are applied, the acceleration is negative and acts in the opposite direction to tht of motion at the instant of braking.

• A body moving with constant speed has zero acceleration only when the particle is in …… dimensional motion.

A body moving with constant speed has zero acceleration only when the particle is in one dimensional motion.

• In velocity-time graph, velocity is taken on ……… axis.

In velocity-time graph, velocity is taken on x-axis

• Area under velocity-time graph gives …………..

Area under velocity-time graph gives the displacement.

• The velocity-time graph of a moving object is a straight line parallel to the time axis. It means the velocity of the object is ………..

The velocity-time graph of a moving object is a straight line parallel to the time axis. It means the velocity of the object is constant.

• The velocity-time graph of a body moving with uniform velocity is a straight line parallel to …………..

The velocity-time graph of a body moving with uniform velocity is a straight line parallel to time axis.

The post Motion in a Straight Line Fill in the Blanks Questions appeared first on The Fact Factor.

In last few articles, we have seen the concept of a motion in a straight line. From this article we shall apply the concept. In this article we shall study true and false type questions based on a motion in a straight line.

State whether the following statements are true or false. If false correct the statement.

• The travel of a train from one station to another is an example of translatory motion.

True

In translational motion every particle of the body has the same displacement.

• Motion of an ant along one of the edge of table is translatory motion.

True

In translational motion every particle of the body has the same displacement.

• The magnitude of displacement can be equal to or lesser than the distance travelled.

True

The shortest distance from the initial position to the final position of the body is called the magnitude of the displacement. Thus in case of body moving in a straight line in the same direction, the maximum displacement can be equal to the distance travelled. In all other case, it will be less than the distance travelled.

• Ddistance covered by a moving body is always greater than zero.

True.

• Displacement of a particle can be less than or greater than or equal to zero.

True

• Uniform speed is a vector quantity

False

Correction: Speed is a scalar quantity

Speed = distance / time, in this formula, both the dstance travelled and time are scalar quantities, hence speed is a scalar quantity.

• A particle moving with a uniform velocity must be along a straight line.

True.

Since velocity is a vector quantity it has both the magnitude and direction and if direction changes it is not uniform velocity. In case of circular motion there is a continuous change in direction leading to accelarated motion which results in non uniform velocity.

• A body can have a constant speed and still have varying velocity.

True. 

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. A body can not have its velocity constant, while its speed varies.

• The magnitude of average velocity is always equal to the average speed.

False.

Correction: The magnitude of average velocity ned not be equal to the average speed.

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. The magnitude of the average velocity of an object is equal to its average speed, only in one condition when an object is moving in a straight line.

• Average velocity can be calculated by taking the average of initial and final velocities for a given time interval irrespective of the type of acceleration.

True

• For a body moving along a circular path, the average velocity and average speed can never be equal.

True.

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. Thus in case of uniform ciercular motion, average speed is constant and equal to the magnitude of the instantaneous velocity of the body, but average velocity is zero.

• Average velocity can be zero, but average speed of a moving body can not be zero in any finite time interval.

True

uniform ciercular motion, average speed is constant and equal to the magnitude of the instantaneous velocity of the body, but average velocity is zero.

• If a body moves with constant velocity, its displacement depends on depends on square of time

False

Correction: If a body moves with constant velocity, its displacement depends on depends on time

v = ds/dt = k = constant

Integrating both sides with time t

s = kt

Thus displacement varies directly with time (t)

• A particle speed is constant, acceleration of the particle must be zero.

False

Correction: A particle speed is constant, acceleration of the particle need not be zero.

In a uniform circular motion, the speed of the body is constant but due to continuous change in direction, the velocity is varying. Thus particle possesses acceleration.

• When a particle moves with a constant speed in the same direction, neither the magnitude nor the direction of velocity changes and so acceleration is zero.

True

Since velocity is a vector quantity it has both the magnitude and direction. In this case both the speed and direction are the same. Hence the particle is moving with a constant velocity and has zero acceleration.

• A particle is known to be at rest at time t = 0. If its acceleration at t = 0 is zero.

False

Correction: A particle is known to be at rest at time t = 0. If its velocity at t = 0 is zero.

A body is said to be at rest if it does not change its position with respect to its immediate surroundings. Thus the velocity of the body decides its state of motion.

• An object covers distances in direct proportion to the square of the time elapsed. Its acceleration is increasing.

False

An object covers distances in direct proportion to the square of the time elapsed. Its acceleration is constant

s = kt² (given)

Differentiating both sides w.r.t. time t

velocity = v = ds/dt = 2kt

Differentiating both sides again w.r.t. time t

acceleration = a = dv/dt = 2k = constant

Thus in this case acceleration is constant. i.e. the object is moving with constant acceleration.

• A particle in one-dimensional motion with a positive value of acceleration must be speeding up.

False

A particle in one-dimensional motion with a positive value of acceleration may or may not be speeding up.

If the initial velocity of a body is negative then even in case of positive acceleration, the body speeds down. 

• There can be a motion in which speed is constant but velocity is variable.

True

In uniform circular motion, speed is constant but velocity is variable.

• A body moves with retardation when it is projected vertically upward.

True

Every body on the earth surface is acted upon by gravitational force acting in downward direction. When a body is projected vertically upward, due to the action of the gravitational force, its velocity goes on decreasing. Thus the body moves with retardation. at the highest point of its journey its velocity is zero.

• A body is projected vertically up. On reaching maximum height, its velocity becomes zero.

True

Every body on the earth surface is acted upon by gravitational force acting in downward direction. When a body is projected vertically upward, due to the action of the gravitational force, its velocity goes on decreasing. Thus the body moves with retardation. at the highest point of its journey its velocity is zero.

• A stone dropped from a height moves with constant velocity

False

Correction: A stone dropped from a height moves with constant acceleration.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

• When two balls of different masses are thrown vertically upwards with the same initial speed, the heavier body rises to greater height then the lighter body.

False

Correction: When two balls of different masses are thrown vertically upwards with the same initial speed, both the bodies will rise to the same maximum height.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity. The height reached by the body epends on the initial speed by which they are thrown ipwards and the acceleration due to gravity at that place. Both the bodies will be acted upon by the same acceleration due to gravity and they are projected with same initial speed. hence both the bodies will rise to the same height.

• The distance travelled by a freely falling body in every successive second is the same.

False:

Correction: The distance travelled by a freely falling body in every successive second increases.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity. Thus freely falling body moves with acceleration due to gravity in downward direction i.e. in the direction of motion, hence its speed increases continuously and thus it covers more distance in every successice second.

• The area under the velocity-time diagram shows the displacement of the body

True

• Velocity-time graph cannot be used to find the instantaneous velocity

False

Correction: Velocity-time graph can be used to find the instantaneous velocity

• Velocity-time graph can be used to find displacement of the body.

True

The area under the velocity-time diagram shows the displacement of the body

• Equations of motion are applicable only when a body moves with uniform velocity.

False

Correction: Equations of motion are applicable only when a body moves with uniform acceleration.

• Direction of motion is decided by the displacement of a body.

False

Direction of motion is decided by the velocity of a body. Positive value of velocity indicates body is moving in the direction of the displacement while neghative value of velocity indicates the body is moving in the opposite direction to that of displacement.

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The post Motion in a Straight Line True and False Questions appeared first on The Fact Factor.

In the last article we have studied the concept of speed and velocity. Motion is an important part of our life. Our daily activities involve motion of different kinds. When we study motion, we come across another important concept namely acceleration. In this article, we shall study the the concept of uniform acceleration.

Notes
Very Short Answer Type Questions
Short Answer Type Questions
Concept Application

Velocity:

The rate of change of displacement of a body with respect to time is called the velocity of the body.

Velocity = Displacement / Time

Velocity is a vector quantity, its S.I. unit is m/s and c.g.s. unit is cm/s. Its dimensions are [L¹M⁰T^-1].

Uniform Velocity: 

When the magnitude and direction of the velocity of a body remain the same at any instant, then the body is said to have uniform velocity. For uniform motion acceleration a = 0 and Displacement = velocity × time.

Example: The velocity of light in a particular medium is uniform velocity.  The velocity of sound in air at constant temperature is uniform velocity.

Non Uniform Velocity: 

When the magnitude of velocity or the direction of velocity or both changes at any instant the body is said to have the nonuniform velocity or variable velocity.

A body can have non-uniform velocity in the following three cases.

• When the direction of the velocity of a body remains the same but its magnitude changes continuously then the body has variable velocity. e.g. a ball is thrown vertically upward.
• When the magnitude of the velocity of a body remains the same but the direction changes continuously then the body has variable velocity. e.g. uniform circular motion of a body.
• When both the magnitude and direction of the velocity of body change continuously, then the body has variable velocity. e.g. ball thrown by making the acute angle with the horizontal (projectile motion)

When a body has variable velocity, then it has acceleration.

Acceleration:

The rate of change of velocity with respect to time is called acceleration.

Acceleration is vector quantity its S.I. unit is m/s². Its dimensions are [L¹M⁰T^-2].

Acceleration = (v – u)/t

Where, u = Initial velocity

v = Final velocity

t = Time in which the change takes place

Acceleration can be positive, negative or zero. If the velocity is increasing then acceleration is positive. If the velocity is decreasing acceleration is negative. If the velocity is the constant acceleration is zero. Negative acceleration is also called deceleration or retardation.

If the velocity is increasing then the direction of acceleration is same as that of the velocity of the body. If the velocity is decreasing then the direction of acceleration is opposite to that of the velocity of the body.

It is to be noted that the velocity and not the acceleration of the body determines the direction of motion.

Uniform Acceleration:

When equal changes take place in the velocity of a body in equal interval of time, then the acceleration is called uniform acceleration. e.g. the motion under gravity.

Variable Acceleration:

When The change in the velocity of a body in equal interval of time is not constant, then the acceleration is called non-uniform acceleration. Example: the motion of a vehicle on crowded road.

Acceleration Due To Gravity:

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

It is denoted by ‘g’. It varies from place to place. The average value of g at sea level is taken as 9.8 ms^-2 in S.I. system and 980 cms^-2 in c.g.s. system. When solving problems on the motion under gravity as per the convention the value of ‘g’ should be negative.

Relation Between Velocity and Acceleration:

• When both velocity and acceleration are positive, acceleration is in the direction of velocity and velocity increases.
• When velocity is positive and acceleration is negative, the acceleration is in the opposite direction of the velocity and velocity decreases.
• When velocity is negative and acceleration is positive, the acceleration is in opposite direction of velocity and velocity increases but body is moving in opposite direction.
• When both velocity and acceleration are negative, acceleration is in the direction of velocity and velocity decreases but body is moving in opposite direction.

Concepts:

Very Short Answer Type Questions

Q1. The average value of acceleration due to gravity at sea level is …… ms^-2

The average value of acceleration due to gravity at sea level is 9.8 ms^-2

Q2. Define acceleration.

The rate of change of velocity with respect to time is called acceleration

Q3. Give c.g.s., m.k.s. and S.I. units of acceleration.

Q4. Retardation is ……….. quantity

Retardation is a vector quantity

Q5. ……. acceleration is called retardation or deceleration.

Negative acceleration is called retardation or deceleration.

Q6. A freely falling body falls with …………..

A freely falling body falls with uniform acceleration, called acceleration due to gravity.

Q7. When is a body said to have zero acceleration?

If a body is at rest or moving with uniform velocity, then the body is said to have zero acceleration.

Q8. What is the acceleration of a body when its velocity remains constant?

In such case the acceleration of the body is zero.

Q9. What is acceleration due to gravity?

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

Q10. Rate of decrease in the magnitude of velocity is called ……..

Rate of decrease in the magnitude of velocity is called retadation or deceleration.

Q11. Give one example of each type of following motions

Uniform acceleration: free fall of a body under influence of gravity

Variable acceleration: A motion of vehicle on crowded road

Q12. What is retardation?

Negative acceleration is called retardation or deceleration.

Q13. Is acceleration due to gravity constant everywhere?

Acceleration due to gravity is not constant everywhere. It depends on altitude, depth, shape of earth, and latitude of the place. It is maximum at the poles and minimum at the equator.

Short Answer Type Questions

Q1. Explain the term ‘acceleration due to gravity’.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity.

It is denoted by ‘g’. It varies from place to place. The average value of g at sea level is taken as 9.8 ms^-2 in S.I. system and 980 cms^-2 in c.g.s. system. When solving problems on the motion under gravity as per the convention the value of ‘g’ should be negative.

Q2. Which of the following bodies does hit the ground first when realeased from the same height simultaneously: a body of mass 1 kg or a body of mass 10 kg.

When the body falls freely under gravity, the acceleration produced in the body due to the gravitational force of attraction of the earth, then the acceleration by which the body falls down is called the acceleration of gravity. The value of g does not depend on mass of the body. If the two bodies of different masses are dropped from the same height simultaneusly, both will reach ground simultaneously, if the effect of air (friction and buoyancy) is neglected.

Q3. Distinguish between acceleration and retardation.

Q4. Distinguish between uniform acceleration and variable acceleration.

Concept Application:

Q1. A train moving with a speed 90 kmph is brought to rest in 10 s. Find its retardation.

u = Initial velocity = 90 kmph = 90 x (5/18) = 25 ms^-1

v = Final velocity = 0

Time in which change is brought = 10 s

We have acceleration = a = (v – u)/t

a = (0 – 25)/10 = – 2.5 ms^-2

Negative sign indicates retardation.

Q2. A car initially at rest attains velocity of 20 ms^-1 with uniform acceleration in 2.5 s. What is its acceleration?

u = Initial velocity = 0

v = Final velocity = 20 ms^-1

Time in which change is brought = 2.5 s

We have acceleration = a = (v – u)/t

a = (20 – 0)/2.5 = 8 ms^-2

Q3. The velocity of an object increases at a constant rate 20 m s^-1 to 50 m s^-1 in 10s. Find the acceleration.

Given: Initial velocity = u = 20 m s^-1, Final velocity = v = 50 m s^-1, Time elapsed = t = 10 s.

To Find: Acceleration = a =?

Solution:

a = (v – u)/t = (50 – 20)/10 = 30/10 = 3 m s^-2

Acceleration of the object is 3 m s^-2

Q4. A stone is thrown vertically upwards with an initial velocity 50 m s^-1 comes to halt in 5 s. Find the acceleration.

Given: Initial velocity = u = 50 m s^-1, Final velocity = v = 0 m s^-1, Time elapsed = t = 5 s.

To Find: Acceleration = a =?

Solution:

a = (v – u)/t = (0 – 50)/5 = – 50/5 = – 10 m s^-2

The negative sign indicates retardation

Retardation of the stone is 10 m s^-2

Q5. The velocity of an object increases at a constant rate 54 km h^-1 to 72 km h^-1 in 5 s. Find the acceleration.

Given: Initial velocity = u = 54 km h^-1= 54 x (5/18) = 15 m s^-1, Final velocity = v = 72 km h^-1= 72 x (5/18) = 20 m s^-1, Time elapsed = t = 5 s.

To Find: Acceleration = a =?

Solution:

a = (v – u)/t = (20 – 15)/5 = 5/5 = 1 m s^-2

Acceleration of the object is 1 m s^-2

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In this article, we shall study the concept of acceleration due to gravity and its characteristics. Also, we shall derive an expression for the same and solve some numerical problems.

Weight of a Body:

Weight of a body is the force with which the body is attracted towards the centre of the earth (planet).

Its unit is newton (N) and dimensions are the same as that of the force [M¹L¹ T^-2]. Mathematically the weight of a body on the surface of the Earth (Planet) is given by

W = F = mg

Where m = mass of the body and

g = acceleration due to gravity on the surface of the earth

Force of Gravitation:

The force of attraction between two material bodies in the universe is known as the force of gravitation. If one of the body is the earth or some other planet or natural satellite then the force of gravitation is called the force of gravity.

Acceleration Due to Gravity:

When a body is released from a height, it gets accelerated towards the earth with constant acceleration, this constant acceleration is called the acceleration due to gravity.

Expression for Acceleration Due to Gravity on the Surface of the Earth (Planet):

Let us consider a body of mass ‘m’ be at rest on the surface of the earth on which the acceleration due to gravity is ‘g’.  If ‘M’ and ‘R’ are mass and radius of the earth (planet) respectively

Now the force of attraction on the body is equal to its weight ‘mg’

This is the expression for acceleration due to gravity on the surface of the earth. Thus acceleration due to gravity on the surface of the earth (planet) is directly proportional to the mass of the earth (planet) and inversely proportional to the square of the radius of the earth (planet).

Characteristics of Acceleration Due to Gravity:

• It is on account of the gravitational force acting on the body.
• Average acceleration due to gravity on the surface is different for different planets.
• At a given place, the value of acceleration due to gravity is the same for all bodies irrespective of their masses.
• It changes from place to place. i.e. it changes with the change in latitude or altitude or depth.
• The acceleration due to gravity at a small height ‘h’ from the surface of the earth is the same as the acceleration due to gravity at the depth, ‘d = 2h’ below the surface of the earth. It means that the value of acceleration due to gravity at a small height from the surface of the earth decreases faster than the value of the acceleration due to gravity at the depth below the surface of the earth.

Notes:

• The value of acceleration due to gravity at the sea level and latitude 45° is taken as the standard.
• The value of acceleration due to gravity at the equator is 9.7804 m/s² and at poles is 9.8322 m/s².
• Unless otherwise stated, the value of ‘g’ is taken as 9.81 m/s² in S.I. system and 981 cm/s².
• The values of ‘g’ for Delhi, Kolkata, and Mumbai are 9.7914 m/s², 9.7876 m/s², 9.7863 m/s² respectively.

Difference Between Universal Gravitation Constant (G) and Acceleration Due to Gravity (g):

• Universal gravitation constant ‘G’ is a scalar quantity while acceleration due to gravity ‘g’ is a vector quantity.
• Universal gravitation constant ‘G’ is universal constant, while acceleration due to gravity ‘g’ changes from place to place and from planet to planet.
• Dimensions of Universal gravitation constant ‘G ‘are [M^-1L³ T^-2], while dimensions of acceleration due to gravity ‘g’ are [M⁰L¹ T^-2].

Expression for acceleration due to gravity At a Height ‘h’ From the Surface of the Earth:

Let us consider a body of mass ‘m’ at rest at height ‘h’ from the surface of the earth. Let the acceleration due to gravity at this height be is ‘g_h’. Let ‘r’ be the distance of the body from the centre of the earth. If ‘M’ and ‘R’ are mass and radius of earth respectively then, r = R +  h

Where r = R + h

This is the expression for the acceleration due to gravity at height h from the surface of the earth.

Relation Between g and g_h:

Numerical Problems on Acceleration Due to Gravity:

Example – 01:

Taking G = 6.67 × 10^-11 N m²/kg², the radius of the earth as 6400 km and mean density of earth as 5500 kg/m³, calculate g at the surface of the earth.

Given: Radius of the Earth = R = 6400 km = 6.4 × 10⁶ m, Density of material of earth = ρ = 5500 kg/m³, G = 6.67 × 10^-11 N m²/kg²

.To find: Acceleration due to gravity = g =?

Solution:

Ans: Acceleration due to gravity = 9.83 m/s².

Example – 02:

At what height will be the acceleration due to the gravity of the earth fall off to one half that at the surface? At what height will the value of g be 8 m/s²? Take radius of earth = 6400 km.

Solution Part – I:

Given: R = 6400 km and g_h = g/2

To find: Height above the surface of the earth = h =?

Now, r = R + h

∴ h = r – R = 9050 – 6400 = 2650 km

Solution Part – II:

Given: R = 6400 km and g_h = 8 m/s²

To find: Height above the surface of the earth = h =?

Now, r = R + h

∴ h = r – R = 7084 – 6400 = 684 km

Ans: At height of 2650 km, the acceleration due to gravity of the earth fall off to one half that at the surface and at a height of 684 km the acceleration due to gravity is 8 m/s².

Example – 03:

How far from the centre of the earth does the acceleration due to gravity reduce by 5 per cent of its value at the surface of the earth? Take the radius of the earth as 6.4 x 10⁶ m.

Given: , R = 6.4 x 10⁶ m  and g_h = g – 5% g = 0.95 g

To find: Distance of point from centre of the earth =r =?

Solution:

Ans: At the distance of 6566 km from the centre of the earth does the acceleration due to gravity reduce by 5 percent of its value at the surface of the earth

Example – 04:

At a certain height above the surface of the earth, the gravitational acceleration is 90 % of its value on the surface of the earth. Determine the height if the radius of the earth is 6400 km.

Solution:

Given: R = 6400 km  and g_h = 90% g = 0.9 g

To find: Height above the surface of the earth =h =?

Now, r = R + h

∴ h = r – R = 6746 – 6400 = 346 km

Ans: At a height of 346 km from the surface of the earth acceleration due to gravity be 90% of the value at the surface of the earth

Example – 05:

At what height above the earth’s surface will the acceleration due to gravity be 4% of the value at the surface of the earth? R= 6400 km.

Given: R = 6400 km  and g_h = 4% g = 0.04 g

To find: Height above the surface of the earth =h =?

Solution:

Now, r = R + h

∴ h = r – R = 32000 – 6400 = 25600 km

Ans: At height of 25600 km from the surface of the earth acceleration due to gravity be 4% of the value at the surface of the earth

Note:

When solving this type of problem, take care of the phrases reduced by and reduced to. For e.g. if the acceleration is reduced by 5 % data is g_h = g – 5% g = 0.95 g and if the acceleration reduces to 5 % data is g_h = 5% g = 0.05 g.

Example – 06:

The mass of the body on the surface of the earth is 100 kg. What will be its mass and weight at an altitude of 1000 km?

Given: Mass of body = 100 kg, R = 6400 km and altitude = h = 1000 km

To find: Mass and weight at altitude of 1000 km =?

Solution:

r = R + h = 6400 + 1000 = 7400 km

Now weight of body at altitude 100 km = W_h = m g_h = 100 x 7.33 = 733 N

The mass is always constant, hence mass at altitude of 100 km = 100 kg

Ans:  At an altitude of 1000 km the mass of the body is 100 kg and its weight is 733 N.

Example – 07:

A body weights 1.8 kg on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of earth

Given: Weight of body on earth = W_E = 1.8 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/9 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P =?

Solution:

Ans: The weight of the body on the surface of the planet is 0.8 kg or 0.8 kg wt.

Example – 08:

A body weights 4.5 kg on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of the earth?

Given: Weight of body on earth = W_E = 4.5 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/9 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P =?

Solution:

Ans: The weight of the body on the surface of the planet is 2 kg or 2 kg wt

Example – 09:

A body weights 3.5 kg wt, on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/7 that of earth and whose radius is half that of earth

Given: , Weight of body on earth = W_E = 1.8 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/7 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P = ?

Solution:

Ans: The weight of the body on the surface of the planet is 2 kg wt.

Example – 10:

The radius of a planet is half that of the earth. The acceleration due to gravity on the planet’s surface is half that on earth’s surface. Find the mass of the planet in terms of mass M of earth

Given: Acceleration due to gravity on planet   = 1/2 Acceleration due to gravity on earth i.e g_P = 1/2 g_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Mass of the planet = M_P =?

Solution:

Ans: The mass of planet in terms of the mass of earth is M/8

Example – 11:

Find the acceleration due to gravity on the surface of the moon. Given that the mass of the moon is 1/80 times that of the earth and the diameter of the moon is 1/4 times that of the earth. g = 9.8 m/s².

Given: Mass of Moon = 1/80 mass of earth i.e M_M = 1/80 M_E, diameter of Moon = 1/4 diameters of earth i.e.  R_M = 1/4 R_E. , acceleration due to gravity on surface of earth = g_E = 9.8 m/s².

To find: acceleration due to gravity on the surface of moon = g_M =?

Solution:

Ans: The acceleration due to gravity on the surface of the moon is 1.96  m/s²,

Example – 12:

A star having a mass 2.5 times that of the sun and collapsed to a size of radius 12 km rotates with a speed of 1.5 rev/s (Extremely compact stars of this kind are called neutron stars. Astronomical objects pulsars belong to this category). Will object placed on its equator remain stuck to its surface due to gravity? Mass of sun is 2 x 10³⁰ kg.

Given: Mass of Star = 2.5 times mass of Sun i.e M_Star = 2.5 M_Sun, Radius of star  = 12 km = 12 x 10³ m, mass of sun = M_Sun =  2 x 10³⁰ kg. Number of revolutions of star = n = 1.5 rev per second.

Solution:

As the gravitational acceleration on the surface of the star is greater than the centripetal acceleration, the weight of the body will be much larger than the centrifugal force acting on the body. Thus body remains stuck to the surface of the star.

Example – 13:

The mass of the Hubble telescope is 11600 kg. What is its weight and mass when it is in an orbit 598 km above the surface of the earth? Mass of earth is 5.98 x 10²⁴ kg, Radius of earth = 6400 km,

Given: Mass of telescope = m = 11600 kg, Mass of earth = M = 5.98 x 10²⁴ kg, Radius of earth = 6400 km, Height of telescope above the surface of earth = h = 598 km, G = 6.67 x 10^-11 N m²/kg²

To find: Weight of telescope = W =?

Solution:

r = R + h = 6400 + 598 = 6998 km = 6.698  x 10⁶ m

The mass is always constant, hence mass at an altitude of 598 km = 11600 kg

Ans: The weight of Hubble telescope in its orbit is 9.447 x 10⁴ N and mass at an altitude of 598 km = 11600 kg

Example – 14:

Find the value of Universal gravitational Constant G from the following data: M = 6 x 10²⁴ kg, R = 6400 km, g = 9.774 m/s²,
Given: M = 6 x 10²⁴ kg, R = 6400 km = 6.4 x 10⁶ m, g = 9.774 m/s²,
To find: G = ?

Solution:

Ans: The value of G is 6.672 x 10^-11 N m²/kg².

Example – 15: 

Assuming the earth to be homogeneous sphere, find the density of material of the earth from the following data.g = 9.8 m/s², G = 6.673 x 10^-11 Nm²/kg² ,    R = 6400 km,

Given: g = 9.8 m/s², Universal gravitational Constant = G = 6.673 x 10^-11 Nm²/kg² ,R = 6400 km = 6.4 x 10⁶ m,

To find:  Density = r = ?

Solution:

Ans: Density of material of the earth = 5483 kg/m³

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