Show that if a Body is taken to height "h",the acceleration is decreased by the factor [tex]\rm{\dfrac{R^2}{(R+h)^2}}}[/tex],where R is the Radius of the earth.
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Show that if a Body is taken to height "h",the acceleration is decreased by the factor [tex]\rm{\dfrac{R^2}{(R+h)^2}}}[/tex],where R is the Radius of the earth.
No Spamming.
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We know that,
Acceleration due (g):
where g= Acceleration due to gravity
G= Universal gravitational constant
M= Mass
R= Radius
Or,
--------(1)
Let g' be the new acceleration due to gravity at earth's surface at height 'h.' Therefore,
------ From (1)
As we know that g is constant for earth i.e 9.81 m/s². Therefore,
Hence proved.