Gravitation

Question:

A planet of mass M and radius R has a satellite of mass m revolving in acircular orbit around it at a height h above the surface of the planet.

During its motion ,the satellite begins to experience a resistive force (dueto cosmic dust) F=kv where k is constant and v is the instanatous velocity ofthe satlellite

1) Find the initial velocity of the planet ?

a)$v_0=\sqrt{\frac{\mathrm{Gm}}{R+h}}$

b)$v_0=\sqrt{\frac{\mathrm{GM}}{R+h}}$

c)$v_0=\sqrt{\frac{\mathrm{GMh}}{R}}$

d) None of these

2) Find the total intial energy of the satellite ?

a) E$=-\frac{\mathrm{GMm}}{2R}$

b)E$=-\frac{\mathrm{GMm}}{2(R+h)}$

c)E$=-\frac{\mathrm{GMm}}{2(R-h)}$

d)E$=-\frac{\mathrm{GMm}}{2(R+2h)}$

3)Find the velocity of the planet when it touches surfaceof the earth ?

a)$v_f=\sqrt{\frac{\mathrm{GM}}{R-h}}$

b)$v_f=\sqrt{\frac{\mathrm{Gm}}{R}}$

c)$v_f=\sqrt{\frac{G(M+m)}{R}}$

d)$v_f=\sqrt{\frac{\mathrm{GM}}{R}}$

4) Find the time taken by the satellite to fall from itsorbit to the surface of the planet?

a)$t=\frac{m}{2k}{\log }_e\left(\frac{h}{R}\right)$

b)$t=\frac{m}{2k}{\log }_e(1+\frac{2h}{R})$

c)t$t=\frac{m}{2k}{\log }_e(1-\frac{h}{R})$

d)$t=\frac{m}{2k}{\log }_e(1+\frac{h}{R})$

Solution:

Orbital radius =R+h

The orbital speed v of the satellite is given by

$\frac{m{v}_0^2}{R+h}=\frac{\mathrm{GMm}}{{(R+h)}^2}$

or

$v_0=\sqrt{\frac{\mathrm{GM}}{R+h}}$

Now total energy in the orbit is given by

E=PE+KE

$E=-\frac{\mathrm{GMm}}{r}+\frac{1}{2}m{v}^2$

Now substituting the initial velocity in this

$E=-\frac{\mathrm{GMm}}{2(R+h)}$

Final Orbital radius =R

$v_f=\sqrt{\frac{\mathrm{GM}}{R}}$

Let t be the time taken by the satellite to fall from its orbit to thesurface of the planet i.e t is the time taken for the speed to change fromv₀ to v_f .if v is the instantanous speed of the satellite,the resistive force acting on it at that instant is given to be

F=kv

or

$m\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{k}{m}\mathrm{dt}$

Integrating we haveş

${\int }_{u{}_{}}^{v}m\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{k}{m}{\int }_0^t\mathrm{dt}$

${\log }_e\frac{v_f}{v_i}=\frac{k}{m}t$

Substituting the values of v_f and v_i ,we get

$t=\frac{m}{k}{\log }_e{\left(\frac{h+R}{R}\right)}^{1/2}$

$t=\frac{m}{2k}{\log }_e(1+\frac{h}{R})$