Question

A spacecraft is in orbit around Jupiter. The radius of the orbit is 2.800 times the radius of Jupiter (which is RJ = 71,490 km). The gravitational field at the surface of Jupiter is 22.00 N/kg. What is the period of the spacecraft's orbit?

Answer 1

T = 1.132 x 10⁴ s

Step-by-step explanation:

given,

Radius of the orbit = 2.8 times radius of Jupiter

Radius of Jupiter, R_J = 71,490 km

gravitational field at the surface of Jupiter, g_J = 22 N/kg

using Kepler's third law equation

`T^2=(4\pi^2)/(GM)\ r^3`

rearranging the equation,

`T^2=(4\pi^2)/((GM)/(r^2))\ r`

we know,

`g_J = (GM)/(r^2)`

now,

`T^2=(4\pi^2)/(g_J)* r`

`T^2=(4\pi^2)/(22)*71490* 10^3`

T² = 1.2829 x 10⁸

T = 1.132 x 10⁴ s

The period of the spacecraft's orbit is equal to 1.132 x 10⁴ s