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An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location, the acceleration due to gravity is what factor times the value of g at the Earth's surface?

Answer :

Answer:0.25 times

Explanation:

Given

Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

where g'=acceleration due to gravity at that point

Distance from center of Earth=R+R=2R

Gravitational Force is given by

[tex]F=\frac{GM_1M_2}{r^2}[/tex]

Force [tex]F=mg'=\frac{GMm}{4R^2}-----1[/tex]

Force on earth surface [tex]F=mg=\frac{GMm}{R^2}------2[/tex]

Divide 1 and 2 we get

[tex]\frac{g'}{g}=\frac{R^2}{4R^2}[/tex]

[tex]g'=\frac{g}{4}[/tex]  

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