Question

Calculate at what distance from the centre a satellite has to orbit Earth (of mass 6 x 10 24 kg) if it period or revolution is to equal Earth’s rotation period of 23h 56m (about 86, 160 seconds.) What is important about such a satellite? c) The mass of the moon is 81 times smaller (7.4074x 10 22 kg) and the period of orbit is 27.3 days. (Convert using 86, 400 sec = 1 day.) What should the radius of a satellite around the moon be if it is to linger above one spot on the Moon always?

BONUS: Give reasonable arguments to indicate that such a Moon-orbiter may not be possible to establish in practice.

Please answer all questions thank you.

Answer 1

A satellite orbit at a radius, where the centripetal force of the planet is equal to the gravitational force of attraction.

GMm/r​​​​​​2 = mr2

GM/r​​​​​​2 = r2

r= (GM/2)1/3

Here, G = 6.67×10-11 Nm​​​​​2/kg​​​​​2

M= 6×1024 kg

T= 86160 s

= 2π/T

= 7.29246×10-5 rad/s

r= 42219158.81 m

= 42219.158 km

This is the orbit, where the satellite remains stationary relative to the earth. This is the radius of geostationary satellites.

c) Mass of moon, M= 7.4074×1022 kg

T= 27.3×86400s

= 2358720 s

=2π/T

=2.6638×10-6 rad/s

r= 88632886.09 m

= 8.863×107 m

This orbit is not possible, since the force exerted at this distance by earth would be greater than the force exerted by the moon when the object is between the earth and moon.