AP EAMCET · PHYSICS · Gravitation
The acceleration due to gravity at a height of 6400 km from the surface of the earth is \(2.5 \mathrm{~ms}^{-2}\). The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth \(=6400 \mathrm{~km}\) )
- A \(1.11 \mathrm{~ms}^{-2}\)
- B \(1.5 \mathrm{~ms}^{-2}\)
- C \(2.22 \mathrm{~ms}^{-2}\)
- D \(1.25 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(A) \(1.11 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{g}_1=2.5 \mathrm{~m} / \mathrm{s}^2, \mathrm{~h}_1=6400 \mathrm{~km}, \mathrm{~h}_2=12800 \mathrm{~km}\) The acceleration due to gravity at height \(h\) is \(g=\frac{G M}{(R+h)^2} \Rightarrow g \propto \frac{1}{(R+h)^2}\)…
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