The time period of a simple pendulum is given by, T=2πgl.
Therefore, T∝g1.
The acceleration due to gravity at height h from the surface of the earth is given by, g′=(R+h)2g
Therefore,
T2T1=g1g2⇒64=R+hR⇒h=2R=3200km
Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4s and 6s respectively. Then the value of h is -
(consider radius of earth RE=6400km and g on earth 10ms−2)
Held on 28 Jul 2022 · Verified 6 Jul 2026.
1200km
1600km
3200km
4800km
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