Work Done Against Gravity

Work Done Against Gravity

Edited By Vishal kumar | Updated on Sep 25, 2024 06:26 PM IST

In our everyday lives, we frequently encounter situations where work is done against gravity, often without even realizing it. Whether lifting groceries into a car trunk or climbing stairs, these common activities involve exerting force to overcome gravity's pull. Understanding the concept of work done against gravity not only enriches our grasp of basic physics but also helps us appreciate the energy involved in our daily tasks. This article explores this fundamental concept, illustrated with real-life scenarios and a solved example to demonstrate how physics applies to everyday experiences. By delving into the mathematics and principles underlying such actions, we gain a clearer insight into the forces at play in our routine movements.

Work Done Against Gravity

The gravitational potential energy at height 'h' from the earth's surface Is given by

Uh=mgR1+hR


So at the surface of the earth put h=0
We get Us=mgR

So if the body of mass m is moved from the surface of the earth to a point at height h from the earth's surface

Then there is a change in its potential energy.

This change in its potential energy is known as work done against gravity to move the body from the earth's surface to height h.


W=ΔU=GMm[1r11r2]


Where W work done
ΔU change in Potential energy
r1,r2 distances
Putting r1=R, and r2=R+h

So

W=ΔU=GMm[1R1R+h]

when ' h ' is not negligible

W=mgh1+hR

when ' h ' is very small

W=mgh1+hR
But h is small compared to Earth's radius

hR0

SoW=mgh
If h=nR then

W=mgRnn+1
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Solved Examples Based on Work Done Against Gravity

Example 1: A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to : (Neglect the effect of atmosphere.

1) 2gR
2) gR
3) gR/2
4) gR(21)

Solution:

Orbital velocity of the satellite


V=GMrr=R+h

r Position of the satellite from the centre of the earth

V Orbital velocity

wherein
The velocity required to put the satellite into its orbit around the earth.
So

v0=GmR+h

For escape velocity
Apply energy conservation

12mve2=GmmR+hve=2GmR+h=2GmR(h<<R)

required increment in the orbital velocity

vev0=2GmRGmR=GmR(21)vev0=gR(21)

Example 2: The gravitational potential energy of a body of mass ' m ' at the earth's surface is mgRe. Its gravitational potential energy at a height Re from the earth's surface will be ( Re is the radius of the earth)
1) 2mgRe
2) 2mgRe
3) 1/2mgRe
4) 1/2mgRe

Solution:

Work done against gravity
W=ΔU=GMm[1r11r2]

W work done
ΔU change in Potential energy
r1,r2 distances
wherein
If the body is moved from the surface of the earth to a point h above the surface of the earth then use the given formula

ΔU=U2U1=mgh1+hRe as h=ReΔU=mgRe2U2(mgRe)=mgRe2U2=mgRe2+(mgRe)=12mgRe

Hence, the answer is the option 4.

Example 3: If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is :
1) 2mgR
2) 12mgR
3) 14mgR
4) mgR

Solution:
Work done against gravity when ' h ' is not negligible -

W=mgh1+hRW= workdone h height above the surface of the earth R Radius of earth

if h=R
if h=RW=12mgR if h=nRW=mgR(nn+1)n times n=1,2,3
So here
F=Gmmx2 at distance x from centre of earth

dw=Gmmx2dx=Gmm2RRdxx2

Potential energy gained =Gmm|1x|

=Gmm×12R

Gain in potential energy =12mR(GmR2)
=12mR(GMR2)=12mgR[g=GMR2]

Gain in potential energy =12mR(GmR2)=12mgR
Hence, the answer is the option (2).

Example 4: If a body of mass m is moved from the earth's surface at height ‘h’ above the earth work is done against gravity. When (h<<<R) where R is the radius of the earth.

1) -mgR

2) -mgh

3) mgR

4) mgh

Solution:

Work done against gravity when ' h ' is very small

W=mgh1+hRW=mgh
W= work done
h height
wherein

hR0

h is small compared to the earth

Work done against gravity
w=mgh1+hRe if h≪≪<RehRe≪<<<<1

hence w=mgh
Hence, the answer is the option 4.

Summary

This is how the work that the gravitational force does may be determined and calculated. To effectively apply the formula in the exercise questions, be aware of how it was developed. Make sure you utilise the units of these physical quantities appropriately as you practise applying this calculation.

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