Question

A satellite moves in elliptical orbit about a planet. Its maximum and minimum velocities of satellites are $3\times {10}^{4}m/s$ and $1\times {10}^{3}m/s$ respectively. What is the minimum distance of satellite from planet is maximum distance if $4\times {10}^{4}km$?

• A. $4\times {10}^{3}km$
• B. $3\times {10}^{3}km$
• C. $4/3\times {10}^{3}km$
• D. $1\times {10}^{3}km$

Answer 1

Answer: (C)

$4/3\times {10}^{3}km$

Maximum distance of satellite $r_{max}=4\times {10}^{4}km$

Velocity of satellite at this distance $v_{min}=1\times {10}^{3}m/s$

Let the minimum distance of satellite be $r_{min}$.

Velocity of satellite at this distance $v_{max}=3\times {10}^{3}m/s$

Since angular momentum of satellite is conserved i.e. $mvr=$ constant

$\therefore$ $v_{max}{r}_{min}=v_{min}{r}_{max}$

$(3\times {10}^3)\times r_{min}=(1\times {10}^3)\times (4\times {10}^4)$

$⟹$ $r_{min}=4\times {10}^{3}km$