Question

The value of escape velocity on a certain planet is 2 kms ^-1. Then the value of orbital speed for a satellite orbiting close to its surface is

• A. 12 kms ^-1
• B. 1 kms ^-1
• C. √2 kms ^-1
• D. 2√2 kms ^-1

Answer 1

The correct option is C

√2 kms ^-1

Explanation for the correct answer:-

Option (C)

Step 1 Given Data,

$v_e=2km{s}^{-1}$

$v_{o=}?$

Step 2 Formula used,

The orbital speed can be written in terms of escape velocity,

$v_o=\frac{v_e}{\sqrt{2}}$

$v_o$ is the orbital speed for a satellite orbiting close to its surface.

$v_e$ is the escape velocity.

Step 3 Calculating the orbital speed for a satellite,

$v_o=\frac{v_e}{\sqrt{2}}$

$v_o=\frac{2}{\sqrt{2}}=\sqrt{2}km{s}^{-1}$

Hence the correct option is C.