Show that the total mechanical energy of a satellite (mass m) orbiting a distance r from the center of the Earth (mass M-e)is:

-(1/2)*(G m M-e)/(r
if U = 0 at r = infinity.
Show that although friction causes the value of E to decrease slowly, the kinetic energy must actually increase if the orbit remains a circle.

I was looking at this for hours, can't quite figure out where the (1/2) comes from . . . Give me a push or the answer.

Ok, I will bite: what is e?

e is energy I believe.

M-e would be mass of earth.

To show that the total mechanical energy of a satellite orbiting a distance r from the center of the Earth is given by -(1/2)*(G*m*M_e)/r, we need to consider the potential energy and kinetic energy of the satellite.

1. Potential Energy (U):
The potential energy of a satellite at a distance r from the center of the Earth is given by the gravitational potential energy formula:

U = -G * (m * M_e) / r

Here, G represents the gravitational constant, m is the mass of the satellite, M_e is the mass of the Earth, and r is the distance between the satellite and the center of the Earth.

2. Kinetic Energy (K):
The kinetic energy of a satellite is given by the formula:

K = (1/2) * m * v^2

where v is the speed of the satellite.

To find the total mechanical energy (E) of the satellite, we sum up the potential energy and kinetic energy:

E = U + K

Substituting the values of U and K:

E = -G * (m * M_e) / r + (1/2) * m * v^2

Now, let's analyze the effects of friction on the satellite's mechanical energy.

When friction is present, it acts against the motion of the satellite. As a result, the satellite loses mechanical energy, causing a decrease in E over time.

However, because the orbit remains a circle, the satellite's speed (v) must be constant. If v remains constant, the kinetic energy (K) also remains constant, even though the total mechanical energy (E) decreases.

To see why the kinetic energy must increase, we can analyze the equation:

E = -G * (m * M_e) / r + (1/2) * m * v^2

If E decreases, and the potential energy term (-G * (m * M_e) / r) remains constant, the only way to keep E constant is if the kinetic energy term ((1/2) * m * v^2) increases.

Therefore, in order for the total mechanical energy to decrease while maintaining a circular orbit, the kinetic energy must increase to compensate for the loss of potential energy.

Now, let's return to the original equation:

E = -G * (m * M_e) / r + (1/2) * m * v^2

The factor of (1/2) in front of the kinetic energy term arises from the relationship between kinetic energy and speed. It comes from the expression (1/2) * m * v^2, which is derived from the work-energy theorem and the definition of kinetic energy.

Thus, the factor of (1/2) in the total mechanical energy equation is a consequence of the mathematical relationship between kinetic energy and speed.