# A 1520 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.93 × 105 km from the planet’s center. Its angular speed at this radius is the same as the rotational speed of...

A 1520 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.93 × 105 km from the planet’s center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satellite is 24 h . What is the force acting on this satellite? Answer in units of N

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A satellite that orbits around a planet is subjected to a force of gravitational attraction of the planet. Over the satellite it must be exercised a force, equal and opposite to the force of gravity, to prevent its fall to the planet. This analysis allows us to find the force of gravity over the satellite.

The centrifugal force is proportional to the tangential speed of the satellite in its orbit, according to the following expression:

Fc = (mv^2)/r

where:

v, is the tangential velocity of the satellite.

r, is the radius of the orbit.

m, is the mass of the satellite.

In the problem we can find the angular velocity (w) of the satellite.

w = 2π/T

Where T = 24 h, It is the period of rotation of the satellite around the planet

The tangential velocity (v) and the angular velocity (w) are related by the following equation:

v = w*r

Then, we can express (v) as follows:

v = (2π*r)/T

Now, substituting into the equation for the centrifugal force, we have:

Fc = mv^2/r = m[(2π*r/T)^2]/r

Fc = (m*4π^2)r/T^2

Fc = (1520)*(4)(9.87)(1.93*10^5)/(86400)^2

Fc = Fg = 1551 N

**The planet exerts on the satellite, a force of 1551 N**.