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Satellite Energy

Part of the Mechanics section of Coddy's Physics journey. Lesson 93 of 97.

On a circular orbit:

K=GMm2r,U=−GMmr,E=K+U=−GMm2rK = \frac{GMm}{2r}, \quad U = -\frac{GMm}{r}, \quad E = K + U = -\frac{GMm}{2r}

So U=−2K\relax U = -2K and E=−K\relax E = -K.

Three energy bars for a satellite in a circular orbit: kinetic energy K above zero, potential energy U below zero and twice as long, and total energy E below zero and as long as K

In a circular orbit U = −2K, so E = K + U = −K.

A higher orbit has ee closer to zero. Raising a satellite takes the energy W=E2−E1\relax W = E_2 - E_1, even though it ends up slower.

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Challenge

A satellite on a circular orbit has a total energy of −12 GJ-12\ \mathrm{GJ} (−1.2×1010 J-1.2 \times 10^{10}\ \mathrm{J}). Its engine raises it to a circular orbit 3 times as far from the planet's centre. Measuring the radii in units of the first orbit (r1=1\relax r_{1} = 1, r2=3\relax r_{2} = 3), find the energy WW the engine adds.

Try it yourself

Physics diagram: orbit, planet, satellite, path

Solve for WW

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