Question

In 2009, NASA's Messenger spacecraft became the second spacecraft to orbit the planet Mercury. The spacecraft orbited at a height of 125 miles above Mercury's surface. You can assume this is a circular orbit. Determine the orbital speed and orbital period of Messenger. (GIVEN: RMercury= 2.44 x 106m; MMercury= 3.30 x 1023kg; 1 mi = 1609 m)

Answer 1

Answer:

Explanation:

height, h = 125 miles = 125 x 1609 m = 201125 m = 0.2 x 10^6 m

Rdaius, R = 2.44 x 10^6 m

Mass, M = 3.3 x 10^23 kg

Let vo be the orbital speed.

The formula for the orbital speed is given by

$v_{o}=\sqrt{\frac{GM}{R+h}}$

where, M is the mass of mercury, R be the radius of mercury

$v_{o}=\sqrt{\frac{6.67\times10^{-11}\3.3\times 10^{23}}{2.64\times 10^{6}}$

vo = 2887.47 m/s

Time period, T = 2π(R+h) / vo

T = 2 x 3.14 x 2.64 x 10^6 / 2887.47

T = 5741.77 seconds

T = 1.6 hours