Question

Asatellite is revolving nearby Earth of radius = R if its velocity is increased √3/2 V where V is orbital speed of satellite then find maximum distance of satellite from the centre of earth​

Answer 1

Explanation:

IS THE ANSWER

( R / R - 4 ) OR NUMERICALLY 1.00062539086929330831769856160100062 km

Answer 2

A satellite is revolving nearby earth of radius = $R_e$ if its velocity is increased $\sqrt{\frac{3}{2}}v$ where v is orbital speed of satellite.

To find : The maximum distance of satellite from the centre of earth.

solution : orbital speed of satellite = v = $\sqrt{\frac{GM}{R_e}}$ .......(1)

for maximum distance of satellite, its velocity will be minimum.

let mass of satellite is m

so, potential energy of system of earth and satellite = $-\frac{GMm}{R_e}$

kinetic energy of satellite = $\frac{1}{2}m\frac{3}{2}v^2$

from conservation of energy theorem,

⇒$-\frac{GMm}{R_e}+\frac{1}{2}m\frac{3}{2}v^2=-\frac{GMm}{R_{max}}+\frac{1}{2}mv_{min}^2$

from equation (1),

⇒$-\frac{GMm}{R_e}+\frac{3}{4}\frac{GMm}{R_e}=-\frac{GMm}{R_{max}}+\frac{1}{2}mv_{min}^2$

⇒$-\frac{GM}{4R_e}=-\frac{GM}{R_{max}}+\frac{1}{2}v_{min}^2$ ........(2)

from conservation of angular momentum,

$L_1=L_2$

⇒$m\sqrt{\frac{3}{2}}vR_e=mV_{min}R_{max}$

⇒$\sqrt{\frac{3}{2}}\sqrt{\frac{GM}{R_e}}R_e=V_{min}R_{max}$

⇒$\frac{1}{R_{max}}\sqrt{\frac{3GMR_e}{2}}=V_{min}$ .......(2) putting it in equation (2) we get ,

⇒$-\frac{GM}{4R_e}=-\frac{GM}{R_{max}}+\frac{1}{2}\frac{1}{R_{max}^2}\frac{3GMR_e}{2}$

after solving we get, $R_{max}=3R_e$

Therefore the maximum distance of satellite is 3 times of radius of the earth.