Question

Mass of earth: 5.972x10^24 kg

radius of earth: 6,378 km

If you are visiting the international space station, which orbits about 350 km above earth. While you are inside the space station, you are in free fall along with everything else in the space station. a.) draw a free body diagram of all the forces acting on you. b.) what is the net force? c.) write down the equation to calculate the force of attraction between you and earth. d.) How are the two forces (the net force and the force of attraction) related to each other? e.) calculate the acceleration of free fall in m/s^2. Is it less than, equal to or greater than g?

Answer 1

Radius of Earth, r_E = 6378 km = 6378000 m

Height above Earth, h = 350 km = 350000 m

If we are visiting the international space station, then its orbit about 350 km above the earth.

While we are inside the space station, we are in free fall along with everything else in the space station.

(b) The net force which will be given as -

F = m g

where, m = mass of person

g = acceleration due to gravity

(c) An expression for the force of attraction between us and earth which will be given as -

F = G M m / r²

where, G = universal gravitational constant

M = mass of Earth

m = mass of person

r = separation distance b/w two objects

(d) Relationship between the two forces (the net force & the force of attraction) which will be given as -

F_net = F_g

m g = G M m / r²

g = G M / r²

(e) The acceleration of free fall which will be given by -

g = G M / r²

where, G = gravitational constant = 6.67 x 10^-11 N.m²/kg²

M = mass of Earth = 5.972 x 10²⁴ kg

r = distance from the center of Earth = r_E + h

then, we get

g = [(6.67 x 10^-11 N.m²/kg²) (5.972 x 10²⁴ kg)] / [(6378000 m) + (350000 m)]²

g = [(6.67 x 10^-11 N.m²/kg²) (5.972 x 10²⁴ kg)] / (6728000 m)²

g = 8.80 m/s²

It is less than the value of "g". We know that, acceleration due to gravity is 9.80 m/s².