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}] $
