Total mechanical energy for circular orbit (10-13)

Question 1 of 6

Question

Total mechanical energy for a satellite in a circular orbit around Earth

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Question

The greater the radius $r$ of the circular orbit, the greater (less negative) the total mechanical energy.

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Review

As we discussed in Section 10-3, the gravitational potential energy is negative. Making the orbital radius $r$ larger makes the gravitational potential energy less negative—that is, closer to zero—which means that the gravitational potential energy increases. The total mechanical energy is the sum of $K$ and $U_{\mathrm{grav}}$ :

$E = K + U_{\mathrm{grav}} = \frac{1}{2}mv^2 + \left( - \frac{Gm_{\mathrm{Earth}}m}{r}\right) = \frac{Gm_{\mathrm{Earth}}m}{2r} + \left( - \frac{Gm_{\mathrm{Earth}}m}{r}\right)$

or: