Get Answers to all your Questions

header-bg qa

A satellite of mass m, initially at rest on the earth, is launched into a circular orbit at a height equal to the radius of the earth. The minimum energy required is

Option: 1

\frac{\sqrt{3}}{4} m g R \\


Option: 2

\frac{1}{2} m g R \\


Option: 3

\frac{1}{4} m g R \\


Option: 4

\frac{3}{4} m g R


Answers (1)

best_answer

\frac{-G M_{e} m}{R}=E_{initial} \\\\ \begin{array}{l} T E=\frac{1}{2} m v^{2}-\frac{g m M}{2 R} \\ \\ \because \frac{m v^{2}}{2 R}=\frac{G M m}{(2 R)^{2}} \\ \\ T E=\frac{-1}{4} \frac{G M m}{R} \end{array}

Difference in energy - 

\frac{-1}{4} \frac{G M m}{R}-\left(-\frac{G M m}{R}\right)=\frac{3}{4} \frac{G M m}{R}

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE