Question

Calculate the orbital velocity and period of revolution of an artificial satellite of earth moving at an altitude 200km. Radius of earth 6400km, Mass of earth 6x1024 kg and G=6.7x10 Nm kg

Answer 1

g=m1m2÷r2

calculate the value it will come

Answer 2

Given that,

Altitude, h = 200 km

Radius of Earth, $r=6400\ km$

Mass of earth, $M=6\times 10^{24}\ kg$

To find,

The velocity and period of revolution of an artificial satellite.

Solution,

The orbital velocity of a satellite is given by :

$v=\sqrt{\dfrac{GM}{R}}$

R = r + h

$v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{6400000+200000}}\\\\v=7786.93\ m/s$

For period of revolution, using Kepler's third law of motion as :

$T^2=\dfrac{4\pi^2}{GM}R^3\\\\T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 6\times 10^{24}}\times (6400000+200000)^3\\\\T=\left(28360537.6002\right)^{\dfrac{1}{2}}\\\\T=5325.46\ s$

So, orbital velocity is 7786.93 m/s and period of revolution is 5325.46 seconds.

Learn more,

Orbital velocity and period of revolution

https://brainly.in/question/11628385