Question

The minimum energy required o launch a m kg satellite from earth's surface in a circular orbit at an altitude of 2R,R is the radius of earth, will be

• A. 3 mgR
• B. 5/6 mgR
• C. 2 mgR
• D. 1/5 mgR

Answer 1

The correct option is B 5/6 mgR

Energy at earth surface$=P{E}_s+K{E}_r$

$=-\frac{GMm}{R}+K{E}_r------\left(1\right)$

Energy in the orbit at altitude of $2RP{E}_o+K{E}_o$

$=\frac{-GMm}{3R}+\frac{1}{2}m{\left(\sqrt{\frac{GM}{3R}}\right)}^2----\left(2\right)$

(1)=(2)

$K{E}_r=-\frac{GMm}{ER}+-\frac{GMm}{R}$

$K{E}_r=\frac{5GMm}{ER}$