Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
0 votes

The ratio of kinetic energy required to raise a satellite upto a height 'h' to the kinetic energy of the satellite at that height (R-radius of earth)

\[(a)\;2h:R \quad (b)\; h:R\quad (c)\;R:2h \quad (d)\;R:h\]
Can you answer this question?

1 Answer

0 votes
The energy needed to raise the satellite to a height 'h' is
$ \Delta v=v_A -v_B$
$\qquad=\large\frac{GMmh}{R^2(1+\frac{h}{R})}\qquad \frac{GM}{R^2}$$=g$
$E_2=$ energy of the satellite
$\qquad=\large\frac{1}{2}$$ mv_0^2\qquad (v_0-$orbital velocity )
r- distance of satellite from center of earth
$\qquad= \large\frac{1}{2} $$m \bigg(\large\frac{GM}{r}\bigg)$
$\qquad= \large\frac{1}{2}$$ m \bigg(\large\frac{GM}{R+h}\bigg)$
$\qquad= \large\frac{1}{2}$$ m \bigg(\large\frac{GM}{R^2}\bigg)\frac{R}{\bigg(1+\Large\frac{h}{R}\bigg)}$
$\therefore \large\frac{E_1}{E_2}$$=\large\frac{2h}{R}$
Hence a is the correct answer.


answered Aug 22, 2013 by meena.p
edited Feb 17, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App