Chapter 10. Total mechanical energy for circular orbit (10-13)

Question

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Question

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{"title":"Total mechanical energy for a satellite in a circular orbit around Earth","description":"Wrong","type":"incorrect","color":"#99CCFF","code":"[{\"shape\":\"poly\",\"coords\":\"82,133\"},{\"shape\":\"rect\",\"coords\":\"1,40,28,75\"}]"} {"title":"Gravitational constant","description":"Correct!","type":"correct","color":"#ffcc00","code":"[{\"shape\":\"rect\",\"coords\":\"118,11,119,13\"},{\"shape\":\"rect\",\"coords\":\"122,3,155,40\"}]"} {"title":"Mass of Earth","description":"Incorrect","type":"incorrect","color":"#333300","code":"[{\"shape\":\"poly\",\"coords\":\"113,132\"},{\"shape\":\"rect\",\"coords\":\"195,33,199,34\"},{\"shape\":\"rect\",\"coords\":\"158,15,196,39\"}]"} {"title":"Mass of satellite","description":"Incorrect","type":"incorrect","color":"#000080","code":"[{\"shape\":\"rect\",\"coords\":\"263,14,285,44\"}]"} {"title":"Radius of the satellite's orbit","description":"Incorrect","type":"incorrect","color":"#333333","code":"[{\"shape\":\"rect\",\"coords\":\"212,82,234,106\"}]"} {"title":"The greater the radius r of the circular orbit, the greater (less negative) the total mechanical energy.","description":"Wrong","type":"incorrect","color":"#800000","code":"[{\"shape\":\"rect\",\"coords\":\"210,80,241,108\"},{\"shape\":\"rect\",\"coords\":\"2,38,35,75\"}]"}

Question

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Question

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Question

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Question

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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: