A satellite orbits the Earth (mass = 5.98 x 1024 kg) once every = 43200 s. At what radius does the satellite orbit?

Answer 1

Let

T = 43200 s be the satellite's period,

M = 5.98×10²⁴ kg be the mass of the Earth,

G = 6.674×10^-11 m³kg^-1s^-2 be the gravitational constant,

R (to be calculated) be the radius of the satellite's orbit,

v (unknown) be the satellite's velocity.

The satellite's centripetal acceleration

v²/R

must equal its gravitational acceleration

GM/R²

The reason is that the centripetal acceleration is caused by gravity.

The velocity, v, is distance divided by time, i.e.,

v = 2πR/T

By setting the two accelerations equal we get

(2πR)²/T²R = GM/R²

R = (GMT²/4π²)^1/3

Before substituting numbers into the above, let's calculate the dimensions:

(m³kg^-1s^-2 kg s²/4π²)^1/3 = m

So we will get the final answer in meters.

Now substitute actual numbers

R = (6.674×10^-11 × 5.98 × 10²⁴ × 43200² / 4π²)^1/3 = 26.62 x 10⁶ m

So the radius of the satellite's orbit is about 26,620 km,

which approximately 20,000 km above the surface of the Earth.