A satellite in an elliptical orbit has a speed of 9.00km/s when it is at its closes approach to the Earth(perigee). The satellite is 7.00x10^6 m from the center of the Earth at this time. When the satellite is at its greatest distance from the center of the Earth (apogee), its speed is 3.66km/s. Find the distance from the satellite to the center of the Earth at apogee. (assume any energy losses are negligible.)

Question

A satellite in an elliptical orbit has a speed of 9.00km/s when it is at its closes approach to the Earth(perigee). The satellite is 7.00x10^6 m from the center of the Earth at this time. When the satellite is at its greatest distance from the center of the Earth (apogee), its speed is 3.66km/s. Find the distance from the satellite to the center of the Earth at apogee. (assume any energy losses are negligible.)

Answer 1

both inside earth

Answer 2

To solve this problem, we can use the conservation of energy and the conservation of angular momentum.

First, let's use the conservation of energy. The total mechanical energy of the satellite is the sum of its kinetic energy and potential energy:

E = KE + PE

In this case, the kinetic energy is given by:

KE = (1/2)mv^2

where m is the mass of the satellite and v is its velocity.

The potential energy can be calculated using the formula:

PE = (-GMm)/r

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the satellite and the center of the Earth.

Since we are assuming any energy losses are negligible, the total mechanical energy of the satellite is constant throughout its orbit. Therefore, we can equate the total mechanical energy at perigee and apogee:

E_perigee = E_apogee

Now, let's find the mechanical energy at perigee:

E_perigee = KE_perigee + PE_perigee

Since the satellite is at its closest approach to the Earth at perigee, its velocity is given as 9.00 km/s, and its distance from the center of the Earth is 7.00 x 10^6 m. The mass of the satellite does not appear in the equation because it cancels out.

KE_perigee = (1/2)mv_perigee^2 = (1/2)m(9.00 x 10^3)^2 = 0.5 x 9.00 x 10^3^2

PE_perigee = (-GMm)/r_perigee = (-GM)/r_perigee

Substituting these values into the equation for the mechanical energy at perigee, we get:

E_perigee = 0.5 x 9.00 x 10^3^2 + (-GM)/r_perigee

Next, let's find the mechanical energy at apogee:

E_apogee = KE_apogee + PE_apogee

Since the satellite is at its greatest distance from the center of the Earth at apogee, its velocity is given as 3.66 km/s. Let's represent the distance from the center of the Earth at apogee as r_apogee, which we need to find.

KE_apogee = (1/2)mv_apogee^2 = (1/2)m(3.66 x 10^3)^2 = 0.5 x 3.66 x 10^3^2

PE_apogee = (-GMm)/r_apogee = (-GM)/r_apogee

Substituting these values into the equation for the mechanical energy at apogee, we get:

E_apogee = 0.5 x 3.66 x 10^3^2 + (-GM)/r_apogee

Now, equating E_perigee and E_apogee:

0.5 x 9.00 x 10^3^2 + (-GM)/r_perigee = 0.5 x 3.66 x 10^3^2 + (-GM)/r_apogee

We can now solve for r_apogee by rearranging the equation:

(-GM)/r_perigee + GM/r_apogee = 0.5 x 3.66 x 10^3^2 - 0.5 x 9.00 x 10^3^2

By simplifying the equation, we get:

GM(1/r_apogee - 1/r_perigee) = 0.5 x 3.66 x 10^3^2 - 0.5 x 9.00 x 10^3^2

Now, we can substitute values for G, M, r_perigee, and the calculated difference between the kinetic energies at apogee and perigee to solve for r_apogee.