Question

A spacecraft is in a circular orbit around the planet Mars at a height of 140 km. A small part of the spacecraft falls off and eventually lands on the surface of the Mars. The small part has a mass of 1.8 kg. During its fall, the small part loses 0.932 MJ of gravitational potential energy. Calculate the gravitational field strength of Mars.

Answer 1

Answer:

3.7 N/kg

Explanation:

gravitational potential energy = m*g*h

therefore 0.932 x 10⁶ J (MJ to J conversion) = 1.8 * g * 140 * 10³ (km to m conversion)

g = (0.932 x 10⁶)/(1.8*140*10³)

g = 3.7 m/sec²

Gravitational field strength = g = 3.7 m/sec² = 3.7 N/kg

( m/sec² multiply numerator and denominator by 'kg' we get (kg*m/sec²)/kg, now kg*m/sec² = Newton (N) thus it becomes N/kg)