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 (mechanical energy) 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.
M-e is just the mass of the Earth.

Centripetal force is supplied by the gravitational force, so

v^2/r = M G/r^2

Compute the kinetic equation from this equation and add it to the potential energy.

To derive the expression for the total mechanical energy of a satellite orbiting a distance r from the center of the Earth, we need to consider the gravitational potential energy and the kinetic energy of the satellite.

1. Gravitational Potential Energy:
The gravitational potential energy of an object of mass m at a distance r from the center of the Earth (mass M_e) is given by:
U = -(G * m * M_e) / r

Here, G is the universal gravitational constant.

2. Kinetic Energy:
The kinetic energy of the satellite is given by:
K = (1/2) * m * v^2
where v is the orbital velocity of the satellite.

In a circular orbit, the magnitude of the gravitational force is equal to the centripetal force:
(G * m * M_e) / r^2 = m * v^2 / r
Simplifying, we get:
v^2 = (G * M_e) / r

Now, let's substitute this value of v^2 into the expression for kinetic energy:
K = (1/2) * m * (G * M_e) / r

3. Total Mechanical Energy:
The total mechanical energy (E) of the satellite in the orbit is the sum of the gravitational potential energy and the kinetic energy:
E = U + K

Substituting the values of U and K, we get:
E = -(G * m * M_e) / r + (1/2) * m * (G * M_e) / r

Factoring out (G * M_e) / r, we have:
E = -(G * M_e * m / r) * (1 - 1/2)

Simplifying the expression (1 - 1/2), we get:
E = -(G * M_e * m / r) * (1/2)

Hence, the expression for the total mechanical energy of a satellite in a circular orbit is:
E = -(1/2) * (G * M_e * m / r)

Now, let's address the second part of the question regarding friction causing a decrease in mechanical energy.

If there is friction acting on the satellite, it will cause a decrease in the total mechanical energy (E) over time. This decrease is due to the conversion of mechanical energy into other forms, such as heat.

However, for the orbit to remain circular, the centripetal force must remain constant. The only way for this to happen is if the velocity of the satellite increases as the distance from the center of the Earth decreases. This means that the kinetic energy (K) of the satellite must increase.

So, even though the total mechanical energy (E) decreases due to friction, the kinetic energy (K) increases in order to maintain a constant circular orbit.

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), let's break it down step by step.

1. The total mechanical energy of the satellite is the sum of its kinetic energy (KE) and its potential energy (PE), given by the equation:
E = KE + PE

2. The kinetic energy of the satellite (KE) is given by:
KE = (1/2) m v^2
where v is the velocity of the satellite.

3. The potential energy of the satellite (PE) is given by the gravitational potential energy formula:
PE = -G M-e m / r
where G is the gravitational constant, M-e is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth to the satellite.

4. Substituting the equations for KE and PE into the total mechanical energy equation, we get:
E = (1/2) m v^2 - G M-e m / r

Now, let's address the question about the origin of the (1/2) term.

The (1/2) term in the total mechanical energy equation arises from the definition of kinetic energy, which is given by (1/2) m v^2. This factor of (1/2) is a result of integrating the expression for kinetic energy over the velocity range from 0 to v. If you are familiar with calculus, you can derive this relationship through the integral of mv dv.

Moving on to the second part of the question, which asks why the kinetic energy must increase if the orbit remains a circle, despite the gradual decrease in total mechanical energy due to friction.

When a satellite is in a circular orbit, its speed remains constant, and thus its kinetic energy remains constant as well. However, due to friction or drag forces acting upon the satellite, the satellite loses mechanical energy over time.

Since the total mechanical energy is the sum of kinetic and potential energy, and the potential energy remains constant for a circular orbit, the decrease in total mechanical energy must be compensated by an increase in kinetic energy to keep the orbit circular. This means that as the satellite loses potential energy, it gains an equal amount of kinetic energy. Hence, the kinetic energy increases to maintain a circular orbit.

I hope this explanation helps you better understand the derivation of the total mechanical energy equation and why the kinetic energy must increase in a circular orbit despite the decrease in total mechanical energy.