Physics: Gravity

Question

Diwata-1 (aka PHL-Microstat-1) was launched to the International Space Station on March 23, 2016. It was the first Philippine microsatellite and the first satellite built and designed by Filipinos. This was released at a height of 400 kilometers above the Earth’s surface. Assuming a circular orbit, (a) how many hours does it take this satellite to make one orbit? (b) How fast (in m/s) is the Diwata-1 moving? USE THE GUESS METHOD.

Answer 1

Kepler's third law gives the orbital period T as:

T = (2 * pi) * sqrt( a^3 / ( G * M ) )

Here,

a = magnitude of semi-major axis = radius of earth + 400km = 6771 E3 m

G = gravitational constant = 6.67E-11

M = mass of central body = 5.97E24 kg

So, plugging in these values,

T = (2 * pi) * sqrt( (6771E3)^3 / (6.67E-11 * 5.97E24) )

T = 5547 s = 1.54 hours

And for the velocity:

V = d / T = 2 pi a / T

V = ( 2 * pi * 6771E3 m ) / 5547 s

V = 7670 m/s