Assume there are two identical simple pendulum clocks. Clock - 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 4 s and 6 s respectively. Then the value of h is -
(consider radius of earth R_E=6400 km and g on earth 10 m / s² )

Solution

Physics sir · 4 steps

The ratio of time periods directly gives (R+h)/R, so read h straight off it.

1. Period and g. For a simple pendulum T = 2π√(L/g). Both clocks are identical, so L is the same and T ∝ 1/√g.
2. g at height h. At height h, g' = gR²/(R+h)². Taking the square root gives √(g/g') = (R+h)/R, with no square left over.

   T₂/T₁ = √(g/g') = (R+h)/R

3. Put in numbers. T₂/T₁ = 6/4 = 3/2, so (R+h)/R = 3/2. That gives h = R/2 = 6400/2 km.
4. So. h = 3200 km, which is option (c). The value of g on earth is not needed.

The idea to keep. Time period goes as 1/√g and g goes as 1/r², so the period grows in direct proportion to the distance from the earth's centre.

Answer from the NTA final answer key. Solution by PrepWiser.