Aspacecraft 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?

T = 1.132 x 10⁴ s

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

rearranging the equation,

we know,

now,

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

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j