Alanding 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

as technology advances, new experiments often expose problems in accepted theories. c

hope i

-lavira: )