Question

# A 400 kg satellite revolves in a circular orbit of radius $$2R_e$$ about the earth; calculate the kinetic energy, potential energy and total energy of the satellite ? Given : $$R_e = 6.4 \times 10^6 \, m$$

Solution

## $${ M }_{ 1 }=6\times { 10 }^{ 24 }kg\\ { M }_{ 2 }=400kg\\ { R }_{ e }=6.4\times { 10 }^{ 6 }m\\ G=6.67\times { 10 }^{ -4 }{ Nm }^{ 2 }/kg\\ \cfrac { G{ M }_{ 1 }{ M }_{ 2 } }{ 2\left( 2{ R }_{ e } \right) } =\cfrac { 6.67\times { 10 }^{ -11 }\times 6\times { 10 }^{ 24 }\times 400 }{ 2\times 12.8\times { 10 }^{ 6 } } =6.25\times { 10 }^{ 9 }J\\ \therefore K.E=6.25\times { 10 }^{ 9 }J\\ P.E=-2KE=-12.5\times { 10 }^{ 9 }J\\ \therefore T.E=P.E+K.E=6.25\times { 10 }^{ 9 }J$$Physics

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