Question

The international space station (ISS) orbits the Earth with an approximately circular orbit 420 km above Earth's surface (6790 km above the center of the Earth). Suppose one evening the ISS passes in front of the full moon - which subtends an angle of 30 arc minutes (0.5 degrees)- in such a way that it exactly bisects the moon. How long will it take to cross in front of the Moon from one end to the other, as viewed by a person on Earth's surface? Assume the Moon is directly overhead when this happens.

Answer 1

Using : Gravitational force = required centripetal force

GMm/r² = mr$\omega$²

$\omega$ = (GM/r³)^0.5

G = gravitational constant = 6.67*10^-11

M = mass of earth = 5.972*10²⁴ kg

r = radius of orbit = (6371+420)km = 6791 km = 6791000 m

putting the values : $\omega$ = (6.67*10^-11*5.972*10²⁴/6791000³)^0.5

$\omega$ = 1.128*10^-3 rad/sec

angle need to pass = 0.5 degree = 0.5*$\pi$/180 rad = 8.7266*10^-3 rad

using equation : angle = $\omega$t

8.7266*10^-3 = 1.128*10^-3t

t = 7.74 sec