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Chapter 14: Problem 34

(a) Does it take more energy to get a satellite up to \(1600 \mathrm{~km}\) above the Earth than to put it in orbit once it is there? (b) What about \(3200 \mathrm{~km} ?(c)\) What about \(4800 \mathrm{~km}\) ? Take the Earth's radius to be \(6400 \mathrm{~km}\).

Short Answer

Expert verified
Upon calculation, the result reveals that: \( (a) \) It takes more energy to get a satellite to 1600 km than to put it in orbit. \( (b) \) For 3200 km, the energy to lift it and to put it in orbit are similar. \( (c) \) For 4800 km, it takes more energy to put the satellite in orbit than to get it there.

Step by step solution

01

Calculate the gravitational potential energy needed to lift the satellite

Gravitational potential energy is given by \( U = m \cdot g \cdot h \)'. Assuming the gravitational acceleration \( g = 9.8 \mathrm{~m/s^2} \)', mass to be \( m = 1 \mathrm{~kg} \)' as a placeholder, and \( h = 1600, 3200, 4800\mathrm{~km} \)', calculate the energy for each height.
02

Calculate the kinetic energy for the satellite in orbit

To find the speed needed for a stable orbit, we set gravitational force equal to centripetal force: \( G \cdot M \cdot m / r^2 = m \cdot v^2 / r \)'. Solving for \( v \)', we find \( v = \sqrt{G \cdot M / r} \)'. Kinetic energy is then given by \( K = 1/2 \cdot m \cdot v^2 \)'. Note that we use \( r = R_{Earth} + h \)', where \( R_{Earth} = 6400 \mathrm{~km} \)'. Plugging in \( M = 5.972 \times 10^{24} \mathrm{~kg} \)', \( G = 6.67430 ×10^{-11} \mathrm{~m^3 kg^{-1} s^{-2}} \)', calculate \( v \)' and \( K \)' for each height.
03

Compare the energies

Now, compare the potential energy \( U \)' and kinetic energy \( K \)' calculated in the previous steps for each of the three heights 1600km, 3200km and 4800km. The energy required to lift the satellite is larger if \( U > K \)', and it takes more energy to put the satellite in orbit when \( K > U \)'.

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Most popular questions from this chapter

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