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At what height from the centre of the Earth the acceleration due to gravity will be th of its \({\frac{1}{4}}^{th}\) value as at the Earth.
We know that,
Answer variants:
G
R
M
E
Acceleration due to gravity \(g\) \(=\) \(\times\) / (\(^2\))
Where,
\(G\) \(=\)
\(M\) \(=\)
\(R\) \(=\)
Let us assume,
Height from the centre of the Earth (\(R'\)) \(=\) \(R + h\)
The acceleration due to gravity at that height (\(g'\)) \(=\) \(\frac{g}{4}\)
Mass of the earth \(=\)
By substituting we get,
\(\frac{g}{g'}=\)
After simplification,
From the centre of the Earth, the object is placed at
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