I got Part A,B and they are correct. I need help with part C. If you can help me with part C i would really appreciate it. A satellite of mass 5800 kg orbits the Earth and has a period of 5800 s .

Question

Part A Determine the radius of its circular orbit. Express your answer using two significant figures. r = 7.0×10^6 m

Part B Determine the magnitude of the Earth's gravitational force on the satellite. Express your answer using two significant figures. F = 4.8×10^4 N

Part C Determine the altitude of the satellite. Express your answer using one significant figure.

h =_____ m

Answer 1

Ok, just in case let me at first show you how to do Part A and B of the problem, so you can check your answers.

Part A.

I am assuming that I know:

M = 6.0 x 10 ²⁴ kg – mass of the Earth,

R = 6.4 x 10 ⁶ m – radius of the Earth

 g₀ = 9.8 m/s² – acceleration due to gravity on the Earth’s surface.

It is also given to us that

 m = 5.8 x 10 ³ kg – mass of the satellite,

T = 5.8 x 10 ³ s – period of revolution of the satellite.

And for calculations we need to know the value of G = 6.67 x 10 ^{– 11} Nm²/kg² – universal gravitational constant.

Because the satellite is in circular motion, its centripetal acceleration a_c = v²/r, where

v – velocity of the satellite and r – radius of rotation measured from the center of the Earth.

The only force acting on the satellite is gravitational force, hence it causes acceleration equal to the acceleration of free fall. Hence a_c = g.

We know that   

                                                                       a_{c =} v²/r.                                         (1)

To find velocity we will recall the formula

                                                          v = 2πr/T,                                                   (2)

Here r – radius of the orbit and T – period of the revolution of the body. Square of the velocity

                                                           v² = 4 π²r²/T²

Putting the value of velocity to the formula of centripetal acceleration (1), we obtain:

                                                            4 π²r²/ (T²r) = g                                        (3)

At any distance from the surface of the Earth acceleration due to gravity

                                                                     g = GM/r²,                                      (4)

where, as we know, r – radius of rotation measured from the center of the Earth.

On the Earth’s surface

                                                       g₀ = GM/R² = 9.8 m/s² ,                     (5)

where R – radius of the Earth and M – its mass.

From formula (5)

                                                                     GM = g₀R².                                    (6)

If I put it into formula (4), I will have

                                                                        g = g₀R²/r²                                   (7)

Let substitute this value into equation (3), then

                                                                     4π²r/ T² = g₀R²/r² 

Or

                                                                       r³ = g₀R² T²/4 π²      

   

To find the radius of the orbit, we take the qubic root from both sides of the equation.

Then

                                                                   r = (g₀R² T²/4 π²)^1/3                     (8)

Hint: To simplify calculations please use the fact that   g₀  ≡ π² = 9.8 . Which means that the numeric values of those 2 quantities are the same. We can cancel them out in the equation, but keep in mind, that g₀  has unit m/s².

So,                                                r = ((6.4 x 10 ⁶)²∙(5.8 x 10 ³)²/4)^1/3 = 7.0 x 10⁶(m)

Answer to C is h = r - R. h = 0.6 x 10 ⁶ m - details in comments