How high is the Manned Orbital Lab above the Earth's surface and what is its kinetic & potential energy in orbit?The Manned Orbital Lab makes 16.0...

The satellite completes 16 revolutions around the Earth in
one day. The centripetal force acting on the satellite is due to the gravitational force
of the Earth. This is equal to G*Me*Ms/r^2. As the satellite is in a constant orbit, it
is equal to (Ms)*r*w^2, where r is the radius, w is the angular velocity and Ms is the
mass of the satellite.

Equating the two, G*Me*Ms/r^2 =
(Ms)r*w^2

The orbital radius of a satellite is given by the
relation:

R^3 = [(T^2* G* Me) / (4* pi^2)
]

G*Me = 398600, T =
24*3600/16

=> R^3 = [(24*3600/16)^2*
398600/(4*pi^2)]

=> R = 6652
km

Therefore the satellite is 6652 km from the center of
the Earth.

The potential energy of the satellite is
G*Me*Ms/r

= 398600*400/6652 = 23968.7
J

The kinetic energy of the satellite is (Potential
Energy)/2 = 11984.3 J