The time period of simple pendulum in earth surface is T. then time period in satellite is

To find the time period of a simple pendulum in a satellite, we need to consider the factors that affect the time period of a pendulum.

The time period (T) of a simple pendulum is generally given by the formula:

T = 2π * √(L/g),

where:
T = time period of the pendulum,
L = length of the pendulum,
g = acceleration due to gravity.

However, in the case of a satellite, the value of "g" is significantly different from that on the surface of the Earth. This is because in orbit, a satellite experiences an apparent weightlessness due to the balance between the gravitational force acting on it and its centripetal force of motion.

For a satellite in a circular orbit, the acceleration due to gravity (g') acting on it is given by:

g' = (G * M) / R^2,

where:
G = gravitational constant,
M = mass of the Earth,
R = distance between the satellite and the center of the Earth.

The time period of a simple pendulum in a satellite can be calculated by using the modified value of "g" (g') in the formula:

T' = 2π * √(L/g').

So, the time period of a simple pendulum in a satellite is T' = 2π * √(L/((G * M) / R^2)).