Question

In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal different information. You will use the both momentum principle and the energy principle in this problem.
A satellite of mass 3500 kg orbits the Earth in a circular orbit of radius of 7.3 106 m (this is above the Earth's atmosphere).The mass of the Earth is 6.0 1024 kg.
What is the magnitude of the gravitational force on the satellite due to the earth?
F= ________N
Using the momentum principle, find the speed of the satellite in orbit. (HINT: Think about the components of (dp^^\->)\/(dt) parallel and perpendicular to p^^\->.)
v = ________ m/s
Using the energy principle, find the minimum amount of work needed to move the satellite from this orbit to a location very far away from the Earth. (You can think of this energy as being supplied by work due to something outside of the system of the Earth and the satellite.)
work= ________J

Answer 1

Answer:

A) F_g = 26284.48 N

B) v = 7404.18 m/s

C) E = 19.19 × 10^(10) J

Explanation:

We are given;

Mass of satellite; m = 3500 kg

Mass of the earth; M = 6 x 10²⁴ Kg

Earth circular orbit radius; R = 7.3 x 10⁶ m

A) Formula for the gravitational force is;

F_g = GmM/r²

Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Plugging in the relevant values, we have;

F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²

F_g = 26284.48 N

B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.

Thus;

GmM/r² = mv²/r

Making v th subject, we have;

v = √(GM/r)

Plugging in the relevant values;

v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))

v = 7404.18 m/s

C) From the energy principle, the minimum amount of work is given by;

E = (GmM/r) - ½mv²

Plugging in the relevant values;

E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)

E = 19.19 × 10^(10) J