Question

A satellite has a mass of 5850 kg and is in a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.15 x 10 to the 6th power m. What is the true weight of the satellite when itis at rest on the planet's surface?

Answer 1

Answer:

 W = 24.28 kN

Explanation:

given,

Mass of satellite = 5850 Kg

height , h = 4.1 x 10⁵ m

Radius of planet = 4.15 x 10⁶ m

Time period = 2 h

                    = 2 x 3600 = 7200 s

Time period of satellite

$T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}$

R is the radius of planet

h is the height of satellite

$T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}$

now calculation of acceleration due to gravity

$g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}$

$g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}$

g = 4.15 m/s²

True weight of satellite

W = m g

W = 5850 x 4.15

W = 24277.5 N

 W = 24.28 kN

True weight of the satellite is   W = 24.28 kN