Question #174312

Acceleration due to gravity at the earth s surface is 9.8ms-2At what height above the earth is the g equal to 9.8/4ms-2


1
Expert's answer
2021-03-23T11:15:46-0400

The acceleration due to gravity at the Earth's surface can be written as follows:


g1=GMR2.g_1=\dfrac{GM}{R^2}.

The acceleration due to gravity at the height hh above the Earth can be written as follows:


g2=GM(R+h)2.g_2=\dfrac{GM}{(R+h)^2}.

Let's divide g1g_1by g2g_2:


g1g2=(R+h)2R2,\dfrac{g_1}{g_2}=\dfrac{(R+h)^2}{R^2},g1g2=(R+h)R,\sqrt{\dfrac{g_1}{g_2}}=\dfrac{(R+h)}{R},h=R(g1g21),h=R(\sqrt{\dfrac{g_1}{g_2}}-1),h=6.37106 m(9.8 ms22.45 ms21)=6.37106 m.h=6.37\cdot10^6\ m\cdot(\sqrt{\dfrac{9.8\ \dfrac{m}{s^2}}{2.45\ \dfrac{m}{s^2}}}-1)=6.37\cdot10^6\ m.

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