Gravitation Ques 32
- An artificial satellite is moving in a circular orbit around the Earth with a speed equal to half the magnitude of escape velocity from the Earth.
(1990, 8M)
(a) Determine the height of the satellite above the earth’s surface.
(b) If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the Earth, find the speed with which it hits the surface of the Earth.
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Answer:
Correct Answer: 32.(a) 6400 km (b) 7.9 km / s
Solution:
Formula:
- (a) Orbital speed of a satellite at distance $r$ from centre of earth,
$ v_{o}=\sqrt{\frac{G M}{r}}=\sqrt{\frac{G M}{R+h}} \cdots(i) $
Given, $\quad v_{o}=\frac{v_{e}}{2}=\frac{\sqrt{2 G M / R}}{2}=\sqrt{\frac{G M}{2 R}} \cdots(ii)$
From Eqs. (i) and (ii), we get
$$ h=R=6400 \mathrm{~km} $$
(b) Decrease in potential energy
$=$ increase in kinetic energy
or $\frac{1}{2} m v^{2}=\Delta U$
$\therefore \quad v=\sqrt{\frac{2(\Delta U)}{m}}$
$$ \begin{aligned} & =\sqrt{\frac{2 \frac{m g h}{1+h / R}}{m}}=\sqrt{g R} \quad(h=R) \\ & =\sqrt{9.8 \times 6400 \times 10^{3}}=7919 \mathrm{~m} / \mathrm{s}=7.9 \mathrm{~km} / \mathrm{s} \end{aligned} $$
BETA
