UPSC 2019 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

Work-energy theorem · Dynamics & Statics · asked 2× in 13 yrs · Read the full method →

Question

The force of attraction of a particle by the earth is inversely proportional to the square of its distance from the earth’s centre. A particle, whose weight on the surface of the earth is $W$ , falls to the surface of the earth from a height $3h$ above it. Show that the magnitude of work done by the earth’s attraction force is $\dfrac{3}{4}hW$ , where $h$ is the radius of the earth.

Technique

Calibrate $k=Wh^2$ from the surface weight; integrate $\int_h^{4h} k/r^2\,dr$ (equivalently use the $-k/r$ potential).

Solution

Step 1 — Force law and calibration

Let $r$ be the distance from the earth’s centre. The attractive force has magnitude

\mathcal F(r)=\frac{k}{r^2}.

At the surface, $r=h$ (earth’s radius), and the force equals the weight $W$ :

W=\frac{k}{h^2}\quad\Longrightarrow\quad k=Wh^2.

Thus $\mathcal F(r)=\dfrac{Wh^2}{r^2}.$

Step 2 — Limits of the fall

The particle starts at height $3h$ above the surface, i.e. at distance

r_{\text{start}}=h+3h=4h,

and ends on the surface at $r_{\text{end}}=h$ .

Step 3 — Work done by the attraction

The force points inward (toward the centre) and the displacement is inward, so the force does positive work. Its magnitude is

\mathcal W=\int_{h}^{4h}\mathcal F(r)\,dr=\int_{h}^{4h}\frac{Wh^2}{r^2}\,dr =Wh^2\left[-\frac1r\right]_{h}^{4h} =Wh^2\left(\frac1h-\frac1{4h}\right).

Then

\mathcal W=Wh^2\cdot\frac{1}{h}\left(1-\frac14\right)=Wh\cdot\frac34=\frac34 hW.

Answer

\boxed{\;\mathcal W=\frac34\,hW.\;}