# How much work is done when a satellite is launched into orbit?

• tuki

## Homework Statement

A Satellite is brought up into a geostationary orbit (altitude 35800km measured from the surface of the earth). Satellite weights 1000.0kg. How much work is required to bring satellite from a surface of the Earth to
geostationary orbit?

## Homework Equations

Newton's law of universal gravitation
$$F = G \frac{m_1m_2}{r^2}$$
Work is defined as:
$$W \int F \cdot s ds$$

## The Attempt at a Solution

I want to compute work done by Newton's law of universal gravitation when
moving satellite from the surface of the Earth to the orbit. It should be done with:

h0 is surface of the Earth (about 6371 km)
h1 is geostationary orbit (about 6371 km + 35800 km = 42171km)
m1 is mass of the satellite (about 1000 kg)
m2 is mass of the Earth (about 5.972E24)

$$W = \int_{h_0}^{h_1} G \frac{m_1 m_2}{r^2} dr = - G \frac{m_1 m_2}{h_1} - (- G \frac{m_1 m_2}{h_0}$$
$$= G m_1 m_2 (\frac{1}{h_1} - \frac{1}{h_0}) \approx 5.3108\cdot 10^{10} \text{ J}$$

However, our textbook suggests that the correct answer would be 5.77E10 Joules. I can't exactly spot out where the mistake is.

Is giving the satellite potential energy the only thing you need to do to put it in geostationary orbit?

Is giving the satellite potential energy the only thing you need to do to put it in geostationary orbit?
Yes you would need the velocity too in order to stay in orbit.

Yes, I can get the correct answer by adding kinetic energy from the speed $$E_{kinetic} = \frac{1}{2} mv^2$$ to the potential energy I already computed. I wonder if it would be possible to derive the work required to accelerate an object to a given speed (when mass is known) starting from Newton's second law? $$F = ma$$ I mean you start from Newton's second law and end up with a formula for kinetic energy?
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