A landing craft with mass 1.21*104 kg is in a circular orbit a distance 5.90*105 m above the surface of a planet. The period of the orbit is 5900 s. The astronauts in the lander measure the diameter of the planet to be 9.80*106 m. The lander sets down at the north pole of the planet.

What is the weight of an astronaut of mass 84.6 kg as he steps out onto the planet's surface?

W = 661.6 N

Explanation:

The weight of a body is the force of attraction of the plant on the body, so we must use the law of gravitational attraction

F = G m M / r²

Where G is the gravitational attraction constant that values 6.67 10-11 N m² / kg², M is the mass of the planet and r is the distance from the center of the planet.

Let's look for the mass of the planet, for this we write Newton's second law for the landing craft

F = m a

Acceleration is centripetal a = v² / r

G m M / r² = m (v² / r)

The ship rotates rapidly (constant velocity module), let's use uniform kinematic relationships

v = d / t

The distance of a circle is

d = 2π r

v = 2π r / t

We replace

G m M / r² = m (4π² r² / t² r)

G M = 4 π² r³ / t²

M = 4π² r³ / G t²

The measured distance r from the center of the plant is

r = R orbit + R planet

r = 5.90 10⁵ + ½ 9.80 10⁶

r = 5.49 10⁶ m

M = 4 π² (5.49 10⁶) ³ / (6.67 10⁻¹¹ (5.900 10³) ²)

M = 6,532 10²¹ / 2,321 10⁺³

M = 2.814 10²⁴ kg

With this data we calculate the astronaut's weight

W = (G M / R²) m

W = (6.67 10⁻¹¹ 2,816 10²⁴ / (4.90 10⁶) 2) 84.6

W = 7.82 84.6

W = 661.57 N