A 4,000 kg satellite is traveling in a circular orbit 200 km above the surface of the Earth.A 3.00 gram marble is dropped inside the satellite.What is the force of gravity on the marble as viewed by the observers on the Earth? (Me = 5.98 $\times$ 10²⁴ kg,Re = 6.37 $\times$ 10⁶ m,G = 6.67 $\times$ 10^-11 N.m²/kg²) A) 0.405 N B) 0.362 N C) 0.277 N D) 0.202 N E) 0.185 N

Question

Quizplus · Accepted Answer

The force of gravity is given by the formula F = G * ((m1*m2)/r^2), where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.

In this case, the marble has a mass of 0.003 kg and the satellite has a mass of 4000 kg. The distance between their centers is the same as the radius of the satellite's orbit, which is the sum of the radius of the Earth (6370 km) and the altitude of the satellite (200 km):

r = 6570 km = 6.57 x 10^6 m

Plugging in these values, we get:

F = 6.67 x 10^-11 * ((0.003 * 4000)/(6.57 x 10^6)^2) 
  = 0.277 N

Therefore, the force of gravity on the marble as viewed by observers on Earth is 0.277 N.

A 4,000 Kg Satellite Is Traveling in a Circular Orbit ×\times×

A 4,000 Kg Satellite Is Traveling in a Circular Orbit $\times$