Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.0k views
in Physics by (46.6k points)
closed by

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is: 

(1) \(\frac{5GmM}{6R}\)

(2) \(\frac{2GmM}{3R}\)

(3) \(\frac{GmM}{2R}\)

(4) \(\frac{GmM}{3R}\) 

1 Answer

+1 vote
by (49.9k points)
selected by
 
Best answer

Correct option is : (1) \(\frac{5GmM}{6R}\) 

Apply energy conservation,

\(U_i + K_i = U_f + K_f\) 

\(\Rightarrow -\frac{GMm}{R} + K_i = -\frac{GMm}{3R} + \frac{1}{2}mv^2\) 

\(\Rightarrow-\frac{GMm}{R} + K_i = -\frac{GMm}{3R} + \frac{1}{2}\times m\times\frac{GM}{3R}\) 

\(\Rightarrow K_i = -\frac{1}{6} \frac{GMm}{R} + \frac{GMm}{R}\) 

\(K_i = \frac{5}{6} \frac{GMm}{R}\) 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...