A satellite moves in a circular orbit around Earth at a speed of 4390 m/s.

(a) Determine the satellite's altitude above the surface of Earth.
m

(b) Determine the period of the satellite's orbit.
h

am I supposed to use rotational equations for this? I don't know what to start out with.

I tried to use V^2/r = GM(earth)/r^2 but I am not getting the correct answer.

V = sqrt(GM/r)

GM = 3.986x10^14

V = 4390

r = 3.986x10^14/4390^2 = 20,682,748.6m = 20,682.7km

Earth radius = 6378km

Orbital altitude h = 20,682.7 - 6378 = 14,394.7km.

The time it takes a satellite to orbit the earth in a circular orbit, its orbital period, can be calculated from

T = 2(Pi)sqrt[r^3/GM]

where T is the orbital period in seconds, Pi = 3.1416, r = the radius of the circular orbit and GM = the Earth's gravitational constant = 6378x10^42.

I'll let you carry out the computation of the period.

To determine the satellite's altitude above the surface of Earth, we can start by using the formula for the centripetal force required to keep an object in circular motion:

F = mv^2/r

Where F is the gravitational force acting on the satellite, m is the satellite's mass, v is its velocity, and r is the distance from the satellite to the center of Earth.

Since we want to calculate the altitude above the Earth's surface, we can express r as the sum of the Earth's radius (R) and the altitude (h):

r = R + h

Now, we can rewrite the centripetal force equation as:

F = (G * M * m) / (R + h)^2

Where G is the gravitational constant (approximately 6.674 × 10^-11 N(m/kg)^2), and M is the mass of Earth (approximately 5.972 × 10^24 kg).

The force of gravity acting on the satellite is given by:

F = (G * M * m) / R^2

Since the gravitational force is responsible for the centripetal force, we can equate these two equations:

(G * M * m) / R^2 = (G * M * m) / (R + h)^2

By simplifying and rearranging the equation, we can solve for h:

(R + h)^2 = R^2
R^2 + 2Rh + h^2 = R^2
2Rh + h^2 = 0
h^2 + 2Rh - 2Rh = 0
h^2 = 2Rh
h = sqrt(2Rh)

Now, we can substitute the known values into the equation. The mean radius of Earth (R) is approximately 6,371 km (or 6,371,000 m).

h = sqrt(2 * 6,371,000 * (6,371,000 + h))

Since h appears on both sides of the equation, this cannot be solved explicitly. However, we can make an estimation by assuming h << R. In that case, we can neglect the h term when compared to R, leading to:

h ≈ sqrt(2 * 6,371,000 * 6,371,000)
h ≈ sqrt(81,287,496,000) ≈ 285,640 m

Therefore, the satellite's altitude above the surface of Earth is approximately 285,640 meters.

Now, let's move on to determining the period of the satellite's orbit. The period (T) of a satellite can be calculated using the formula:

T = (2πr) / v

Where r is again the distance between the satellite and the center of Earth.

Since r is the sum of the Earth's radius (R) and the altitude (h), we can rewrite the formula as:

T = (2π(R + h)) / v

Now, we can substitute the known values into the formula:

T = (2π(6,371,000 + 285,640)) / 4,390

By evaluating this expression, we can find the time taken for one complete orbit of the satellite around Earth.