Question

A satellite is projected with a velocity $\sqrt{1.5}$ times its orbital velocity just above the earth's atmosphere. The initial velocity of the satellite is parallel to the surface. The maximum distance of the satellite from the earth will be:

• A. 2R
• B. 8R
• C. 4R
• D. 3R

Answer 1

The correct option is D 3R

According to law of conservation of energy
$\frac{-GMm}{R}+\frac{1}{2}m{V}^2=\frac{-GMm}{R+h}$
$=\frac{-GMm}{R}+\frac{1}{2}m(\sqrt{1.5}\sqrt{\frac{GM}{R}}{)}^2=\frac{-GMm}{R+h}$
$=\frac{-GMm}{R}+\frac{3}{4}\frac{GMm}{R}=\frac{-GMm}{R+h}$
$\Rightarrow h=3R$