Gravitation 3 Question 9
12. 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: 12. (a) $-\pi \times 10^{4} km / h \quad$ (b) $3 \times 10^{-4} rad / s$
Solution:
- (a) Orbital speed of a satellite at distance $r$ from centre of earth,
$$ v _o=\sqrt{\frac{G M}{r}}=\sqrt{\frac{G M}{R+h}} $$
Given, $\quad v _o=\frac{v _e}{2}=\frac{\sqrt{2 G M / R}}{2}=\sqrt{\frac{G M}{2 R}}$
From Eqs. (i) and (ii), we get
$$ h=R=6400 km $$
(b) Decrease in potential energy
$=$ increase in kinetic energy
or $\frac{1}{2} m v^{2}=\Delta U$
$\therefore \quad v=\sqrt{\frac{2(\Delta U)}{m}}$
$$ \begin{aligned} & =\sqrt{\frac{2 \frac{m g h}{1+h / R}}{m}}=\sqrt{g R} \quad(h=R) \\ & =\sqrt{9.8 \times 6400 \times 10^{3}}=7919 m / s=7.9 km / s \end{aligned} $$
