Question

A satellite is in orbit 3.117106 m from the center of Earth. The mass of Earth is 5.9821024 kg. Calculate the orbital

period of the satellite.

Answer 1

T = 1733.16 s = 28.88 min

Step-by-step explanation:

The orbital velocity of a satellite about Earth is given as follows:

`v = \sqrt{(GM)/(R)}`

where,

v = orbital speed = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

M = Mass of Earth = 5.982 x 10²⁴ kg

R = Orbit Radius = 3.117 x 10⁶ m

Therefore,

`v = \sqrt{((6.67\ x\ 10^(-11)\ Nm^(2)/kg^(2))(5.982\ x\ 10^(24)\ kg))/((3.117\ x\ 10^(6)\ m))}\\\\v = 11.3\ x\ 10^(3)\ m/s`

but the velocity is given as:

`v = (distance)/(time)`

for distance = circumference = 2πR

time = time period = T = ?

Therefore,

`11.3\ x\ 10^(3)\ m/s = (2\pi(3.117\ x\ 10^(6)\ m))/(T)\\\\T = (2\pi(3.117\ x\ 10^(6)\ m))/(11.3\ x\ 10^(3)\ m/s)\\\\`

T = 1733.16 s = 28.88 min