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# Gravity (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

## Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to gravity. The questions are categorized in terms of the characterizing features of the subject matter :

• Acceleration at a Height
• Acceleration at a Depth
• Comparison of acceleration due to gravity
• Rotation of Earth
• Comparison of gravitational acceleration
• Rate of change of gravity

## Acceleration at a Height

Problem 1 : At what height from the surface of Earth will the acceleration due to gravity is reduced by 36 % from the value at the surface. Take, R = 6400 km.

Solution : The acceleration due to gravity decreases as we go vertically up from the surface. The reduction of acceleration by 36 % means that the height involved is significant. As such, we can not use the approximated expression of the effective accelerations for h<< R as given by :

g = g 1 2 h R g = g 1 2 h R

Instead, we should use the relation,

g = g 1 + h R 2 g = g 1 + h R 2

Note that we have considered reference gravitational acceleration equal to acceleration on the surface. Now, it is given that :

g = 0.64 g g = 0.64 g

Hence,

0.64 g = g 1 + h R 2 0.64 g = g 1 + h R 2

1 + h R 2 X 0.64 = 1 1 + h R 2 X 0.64 = 1

1 + h R = 10 8 = 5 4 1 + h R = 10 8 = 5 4

h R = 5 4 1 = 1 4 h R = 5 4 1 = 1 4

h = R 4 = 6400 4 = 1600 k m h = R 4 = 6400 4 = 1600 k m

Note : If we calculate, considering h << R, then

0.64 g = g 1 2 h R 0.64 g = g 1 2 h R

0.64 R = R 2 h 0.64 R = R 2 h

h = R 1 0.64 2 = 0.18 R = 0.18 X 6400 = 1152 k m h = R 1 0.64 2 = 0.18 R = 0.18 X 6400 = 1152 k m

## Acceleration at a Depth

Problem 2 : Assuming Earth to be uniform sphere, how much a weight of 200 N would weigh half way from the center of Earth.

Solution : Assuming, g = g 0 g = g 0 , the accelerations at the surface (g) and at a depth (g') are related as :

g = g 1 d R g = g 1 d R

In this case,

d = R R 2 = R 2 d = R R 2 = R 2

Putting in the equation of effective acceleration, we have :

g = g 1 R 2 R = g 2 g = g 1 R 2 R = g 2

The weight on the surface corresponds to “mg” and its weight corresponds to “mg’”. Hence,

m g = m g 2 = 200 2 = 100 N m g = m g 2 = 200 2 = 100 N

## Comparison of acceleration due to gravity

Problem 3 : Find the ratio of acceleration due to gravity at a depth “h” and at a height “h” from Earth’s surface. Consider h >> R, where “R” is the radius of Earth.

Solution : The acceleration due to gravity at appoint “h” below Earth’s surface is given as :

g 1 = g 0 1 h R g 1 = g 0 1 h R

The acceleration due to gravity at a point “h” above Earth’s surface is given as :

g 2 = g 0 1 + h R 2 g 2 = g 0 1 + h R 2

Note that we have not incorporated approximation for h>>R. We shall affect the same after getting the expression for the ratio .

The required ratio without approximation is :

g 1 g 2 = g 0 1 h R 1 + h R 2 g 0 g 1 g 2 = g 0 1 h R 1 + h R 2 g 0

g 1 g 2 = 1 h R 1 + h R 2 g 1 g 2 = 1 h R 1 + h R 2

g 1 g 2 = 1 h R 1 + h 2 R 2 + 2 h R g 1 g 2 = 1 h R 1 + h 2 R 2 + 2 h R

For h >> R, we can neglect terms of higher power than 1,

g 1 g 2 = 1 h R 1 + 2 h R g 1 g 2 = 1 h R 1 + 2 h R

g 1 g 2 = 1 h R + 2 h R 2 h 2 R 2 g 1 g 2 = 1 h R + 2 h R 2 h 2 R 2

Again, neglecting term with higher power,

g 1 g 2 = 1 + h R g 1 g 2 = 1 + h R

## Rotation of Earth

Problem 4 : If “ρ” be the uniform density of a spherical planet, then find the shortest possible period of rotation of the planet about its axis of rotation.

Solution : A planet needs to hold material it is composed. We have seen that centripetal force required for a particle on the surface is maximum at the equator. Therefore, gravitational pull of the planet should be as least sufficient enough to hold the particle at the equator. Corresponding maximum angular speed corresponding to this condition is obtained as :

G M m R 2 = m ω 2 R G M m R 2 = m ω 2 R

Time period is related to angular speed as :

ω = 2 π T ω = 2 π T

Substituting for angular speed in force equation, we get the expression involving shortest time period :

T 2 = 4 π 2 R 3 G M T 2 = 4 π 2 R 3 G M

The mass of the spherical planet of uniform density is :

M = 4 π ρ R 3 3 M = 4 π ρ R 3 3

Putting in the equation of time period,

T 2 = 4 x 3 π 2 R 3 G x 4 π ρ R 3 = 3 π G ρ T 2 = 4 x 3 π 2 R 3 G x 4 π ρ R 3 = 3 π G ρ

T = 3 π G ρ T = 3 π G ρ

Problem 5 : Considering Earth to be a sphere of uniform density, what should be the time period of its rotation about its own axis so that acceleration due to gravity at the equator becomes zero. Take g = 10 m / s 2 m / s 2 and R = 6400 km.

Solution : We know that the measurement of gravitational acceleration due to gravity is affected by rotation of Earth. Let g’ be the effective acceleration and g 0 = g g 0 = g . Then,

g = g R ω 2 cos Φ g = g R ω 2 cos Φ

where Φ is latitude angle.

Here , Φ = 0 0, cos Φ = cos 0 0 = 1, Here , Φ = 0 0, cos Φ = cos 0 0 = 1,

g = g R ω 2 g = g R ω 2

Now, angular velocity is connected to time period as :

ω = 2 π T ω = 2 π T

Combining two equations, we have :

g = g R X 4 π 2 T 2 g = g R X 4 π 2 T 2

4 π 2 R = g g T 2 4 π 2 R = g g T 2

T = 4 π 2 R g g T = 4 π 2 R g g

According to question, effective acceleration is zero,

g = 0 g = 0

Hence,

T = 2 π 6400 X 10 3 10 T = 2 π 6400 X 10 3 10

T = π X 1600 s T = π X 1600 s

T = 1.4 h r T = 1.4 h r

## Comparison of gravitational acceleration

Problem 6 : A planet has 8 times the mass and average density that of Earth. Find acceleration on the surface of planet, considering both bodies spherical in shape. Take acceleration on the surface of Earth as 10 m / s 2 m / s 2 .

Solution : Let subscript “1” and “2” denote Earth and planet respectively. Then, ratio of accelerations is :

g 2 g 1 = G M 2 R 2 2 G M 1 R 1 2 = M 2 R 1 2 M 1 R 2 2 g 2 g 1 = G M 2 R 2 2 G M 1 R 1 2 = M 2 R 1 2 M 1 R 2 2

Here,

M 2 = 8 M 1 M 2 = 8 M 1

g 2 g 1 = 8 M 1 R 1 2 M 1 R 2 2 = 8 R 1 2 R 2 2 g 2 g 1 = 8 M 1 R 1 2 M 1 R 2 2 = 8 R 1 2 R 2 2

We need to relate radii in order to evaluate the ratio as above. For this, we shall use given information about density. Here,

ρ 2 = 8 ρ 1 ρ 2 = 8 ρ 1

M 2 V 2 = 8 M 1 V 1 M 2 V 2 = 8 M 1 V 1

But , M 2 = 8 M 1, But , M 2 = 8 M 1,

8 M 1 V 2 = 8 M 1 V 1 8 M 1 V 2 = 8 M 1 V 1

V 1 = V 2 V 1 = V 2

R 1 = R 2 R 1 = R 2

Now, evaluating the ratio of accelerations, we have :

g 2 = 8 g 1 = 8 X 10 = 80 m / s 2 g 2 = 8 g 1 = 8 X 10 = 80 m / s 2

## Rate of change of gravity

Problem 7 : Find the rate of change of weight with respect height “h” near Earth’s surface.

Solution : According to question, we are required to find the rate of change of the weight near Earth’s surface. Hence, we shall use the expression for h<<R/ Also let g 0 = g g 0 = g . Then,

g = g 1 2 h R g = g 1 2 h R

Weight at height, “h”, is given by :

W = m g = m g 1 2 h R = m g 2 m g h R W = m g = m g 1 2 h R = m g 2 m g h R

The rate of change of acceleration due to gravity at a height “h” is given as :

W h = h m g 2 m g h R W h = h m g 2 m g h R

W h = 2 m g R W h = 2 m g R

Problem 8 : What is fractional change in gravitational acceleration at a height “h” near the surface of Earth.

Solution : The fractional change of a quantity “x” is defined as “Δx/x”. Hence, fractional change in gravitational acceleration is “Δ g/g”. Let g 0 = g g 0 = g . Now, effective acceleration at a height “h” near Earth’s surface is given by :

g = g 1 2 h R g = g 1 2 h R

g g = - 2 h g R g g = - 2 h g R

g g g = - 2 h R g g g = - 2 h R

Δ g g = - 2 h R Δ g g = - 2 h R

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