Question

- Calculate the period of a satellite orbiting the Moon, 120 km above the
Moon’s surface. Ignore effects of the Earth. The radius of the Moon is
1740 km and MMoon = 7.348*1022 kg.

Answer 1

356.................

Answer 2

Given:

Height of the satellite from the surface of the moon = 120 km

Radius of the moon = 1740 km

Mass of the moon = 7.348×10²² kg

To find:

Time period of the satellite.

Solution:

Total distance of the satellite from the center of the moon (R) = 120 + 1740 km = 1860 km = 1860* 10³ m

Time period = T = 2π√(R³/MG)

Where M is the mass of the moon

And G is the universal gravitational constant

= 2* 3.14 * √(1860³*(10³)³/ 7.348*10²²*6.67*10⁻¹¹)

=6.28*√131293689*10⁻²

= 6.28* 36234.471 * 10⁻²

= 227552.478 * 10⁻²

=2275.5 seconds (Rounding off to one decimal place)

Therefore the time period is 2275.5 seconds or 37.9 minutes.