Question

Use the universal law of gravitation to solve the following problem.

Hint: mass of the Earth is = 5.97 x 1024 kg


A scientific satellite of mass 1300 kg orbits Earth 200 km above its surface. If Earth has a radius of 6378 km, what is the force of gravity acting on the scientific satellite?


a. Write out the formula for this problem.


b. Plug in the values from this problem into the formula.


c. Solve the problem, writing out each step.


d. Correct answer

Answer 1

a.

`F=G\cdot(M \cdot m)/(r^(2))`

b.

`F=6.67430 * 10^(-11) (N \cdot m^2)/(kg^2) * (5.97 * 10^(24) \ kg * 1,300 \ kg)/((6,578 \ m)^(2))`

c.

`F=6.67430 * 10^(-11) (N \cdot m^2)/(kg^2) * (5.97 * 10^(24) \ kg * 1,300 \ kg)/((6,578 \ m)^(2)) \approx 1.144 * 10^(13) \ N`

d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N

Step-by-step explanation:

a. The formula for finding the force of gravity, F, acting object on an object is given as follows;

`F=G\cdot(M \cdot m)/(r^(2))`

Where;

F = The force acting between the Earth and the object

G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M = The mass of the Earth = 5.97 × 10²⁴ kg

m = The mass of the object

r = The distance between the center of the Earth and the object

b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;

The given mass of the satellite, m = 1,300 kg

The distance between the center of the Earth and the center of the satellite, r = The length of the radius of the Earth + The height of orbit of the satellite

The given height of orbit of the satellite, h = 200 km

∴ r = R + h = 6,378 km + 200 km = 6,578 m

Therefore, by plugging in the values, we get;

`F=6.67430 * 10^(-11) (N \cdot m^2)/(kg^2) * (5.97 * 10^(24) \ kg * 1,300 \ kg)/((6,578 \ m)^(2))`

c. Solving the above equation gives;

`F=6.67430 * 10^(-11) (N \cdot m^2)/(kg^2) * (5.97 * 10^(24) \ kg * 1,300 \ kg)/((6,578 \ m)^(2)) \approx 1.144 * 10^(13) \ N`

d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton