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Acceleration due to gravity is represented as g.The value of g on the surface of earth is 9.8ms−2
The formula for g is g=GMR2
Where G is the universal gravitational constant. Its value = 6.67408×10−11m3kg−1s−2 .M is the mass of earth.R is the radius of earth.
As we see can from the formula, that g∝1R2 -------(1)
Lets take acceleration due to gravity at a height H from the surface as g'
g'∝1(R+H)2------(2)
Lets divide equation 2 with equation 1, to get the relation between g' and g
g'g=R2(R+H)2
⇒g'=g(1+HR)2
Now that we have derived a relation between g'and g, lets solve your question by putting H=2Rin the above equation.
We get g'=g9
⇒ acceleration due to gravity at height 2R form the surface of earth is 19th of g
Note:The above formula is only valid for H>R.
For HR, the relation between is g' and g is
g'=g(1−HR)
If the derivation is required, then feel free to ask, and I will add it.
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