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Question: A small body is initially at a distance \[r\]from the center of earth. \[r\] is greater than the rad...

A small body is initially at a distance rrfrom the center of earth. rr is greater than the radius of the earth. If it takes WWjoule of work to move the body from this position to another position at a distance 2r2r measured from the center of the earth, then how many joules would be required to move it from this position to a new position at a distance of 3r3r from the center of the earth.
A.W5  A.\dfrac{W}{5}\;
B.W3B.\dfrac{W}{3}
C.W2C.\dfrac{W}{2}
D.W6D.\dfrac{W}{6}

Explanation

Solution

To solve the problem the potential energy has to be found and after this, the work done by the body can be calculated using the relation between potential energy and the work done. Note that, the work done has to be calculated for two cases given in the problem and then compared to each other. Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it can be said that gravitational potential energy is an energy that is related to gravitational force or gravity.

Formula used:
Potential energy for the distance R1{R_1} from the center of the earth, U1=−GMmR1{U_1} = - \dfrac{{GMm}}{{{R_1}}}
Potential energy for the distance R2{R_2} from the center of the earth U2=−GMmR2{U_2} = - \dfrac{{GMm}}{{{R_2}}}
And,
Work done W=U2−U1W = {U_2} - {U_1}
G=G = gravitational constant
M=M = Mass of earth
m=m = mass of planet

Complete answer:
Work done by an object in the field of gravitation is defined by the change in potential energy of the object.
Now, for the distance R1{R_1} from the center of the earth, the Potential energy will be U1=−GMmR1{U_1} = - \dfrac{{GMm}}{{{R_1}}}
And, that for the distance R2{R_2} from the center of the earth will be U2=−GMmR2{U_2} = - \dfrac{{GMm}}{{{R_2}}}
G=G = gravitational constant
M=M = Mass of earth
m=m = mass of planet
Hence, by definition the Work done W=U2−U1W = {U_2} - {U_1}
For the 1st case,
Given that, R1=r{R_1} = rand, R2=2r{R_2} = 2r and work done = WW
So, W=−GMm2r−(−GMmr)W = - \dfrac{{GMm}}{{2r}} - \left( { - \dfrac{{GMm}}{r}} \right)
⇒W=GMm2r\Rightarrow W = \dfrac{{GMm}}{{2r}}
Now in the 2nd case, the body moves from a distance R1=2r{R_1} = 2rto a distance R2=3r{R_2} = 3r
If work done is W′W'
then, W′=−GMm3r−(−GMm2r)W' = - \dfrac{{GMm}}{{3r}} - \left( { - \dfrac{{GMm}}{{2r}}} \right)
⇒W′=13GMm2r\Rightarrow W' = \dfrac{1}{3}\dfrac{{GMm}}{{2r}}
⇒W′=13W\Rightarrow W' = \dfrac{1}{3}W [since, ⇒W=GMm2r \Rightarrow W = \dfrac{{GMm}}{{2r}}]

Option B.⇒W3B. \Rightarrow \dfrac{W}{3} is correct.

Note:
The force of gravity can be considered as a conservation force since the work done does not depend upon the path followed. The potential energy that arises due to this force is called gravitational potential energy. Also, Newton’s law of universal gravitation states that the gravitational force attraction between any two particles of different masses separated by distance will have magnitude F. This is gravity force and here we get a new constant called gravitational constant.