Question

How much time a satellite in an orbit at height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?

Hint:

Remember: The orbital time period is inversely proportional to the square root of the mass of the central body. Increasing the mass of the earth reduces the time period.

Answer 1

The time period T of a satellite in orbit is given by the formula: T = 2 π √((r³)/(GM)) Where:
- T is the orbital time period,
- r is the distance from the center of the earth,
- G is the universal gravitational constant,
- M is the mass of the earth.
Given that the height of the satellite above the earth’s surface is 35780 km, the total distance from the center of the earth is: r = 35780 km + 6371 km = 42151 km = 4.2151 × 10⁷ m If the mass of the earth is increased by a factor of 4, the new mass M' becomes: M' = 4M Substituting the new mass into the orbital time formula, we get: T' = 2 π √((r³)/(G × 4M)) = (T)/(2) Thus, the new orbital time period will be half of the original time period, i.e., the satellite will complete its orbit in half the time.