Gravitation Question 19

Question: The value of g on the earth’s surface is $ [980,cm/{{\sec }^{2}}] $ . Its value at a height of 64 km from the earth’s surface is [MP PMT 1995]

Options:

A) $ [960.40,cm/{{\sec }^{2}}] $

B) $ [984.90,cm/{{\sec }^{2}}] $

C) $ [982.45,cm/{{\sec }^{2}}] $

D) $ [977.55,cm/{{\sec }^{2}}] $

Show Answer

Answer:

Correct Answer: A

Solution:

$ [\frac{g’}{g}={{\left( \frac{R}{R+h} \right)}^{2}}={{\left( \frac{6400}{6400+64} \right)}^{2}}] $ $ [\Rightarrow ] $ $ [g’=960.40\ cm/s^{2}] $



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