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Numericals · Q47

Q.What is the gravitational potential due to the Earth at a point which is at a height of 2RE2R_E above the surface of the Earth? Mass of the Earth is 6×10246\times10^{24} kg, radius of the Earth = 6400 km and G=6.67×10−11G=6.67\times10^{-11} N m^2 kg^-2.

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At a height of 2RE2R_E above the surface, the distance from the Earth's centre is r=RE+2RE=3REr=R_E+2R_E=3R_E. Gravitational potential is V=−GMr=−GM3RE.V=-\dfrac{GM}{r}=-\dfrac{GM}{3R_E}. Given M=6×1024M=6\times10^{24} kg, RE=6400R_E=6400 km =6.4×106=6.4\times10^6 m, G=6.67×10−11G=6.67\times10^{-11} N m^2/kg^2:

GM=6.67×10−11×6×1024=4.002×1014GM=6.67\times10^{-11}\times6\times10^{24}=4.002\times10^{14} m^3/s^2.

r=3×6.4×106=1.92×107r=3\times6.4\times10^6=1.92\times10^7 m. …

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