Gravitation and Circular Orbits

[kevi555]

Just a question about gravitation equations:

Neglecting Earth's rotation, show that the energy needed to launch a satellite of mass m into circular orbit at altitude h is equal to:

$$(\frac {GMm}{R})(\frac{R+2h}{2(R+h)})$$

Where R = the radius of the Earth and M = the mass of the Earth.

Any help would be truly appreciated! Thanks all.

 

[G01]

This sounds like a homework problem. Have you put any thought into this problem? If you have tell us what you tried. You need to show some work to get help with homework type problems.Also, for homework type questions, please post them in the homework sections, not the science forums, since these are meant more for discussion in the topic area and non homework type problems (conceptual questions and questions that do not seem to come from a textbook).

 

[Moetasim]

Hello,

Thank you for your question about gravitation equations. The equation you have provided is known as the vis-viva equation, which relates the energy of a satellite in circular orbit to its distance from the center of the Earth.

To understand why this equation is valid, we first need to look at the concept of gravitational potential energy. This is the energy that an object possesses due to its position in a gravitational field. The closer an object is to the center of gravity, the more potential energy it has.

In this case, we can calculate the potential energy of a satellite of mass m at a distance R+h from the center of the Earth as:

U = -\frac{GMm}{R+h}

Where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth. This equation takes into account the effect of Earth's mass on the satellite's potential energy.

Now, let's consider the kinetic energy of the satellite, which is the energy it possesses due to its motion. In circular motion, the satellite is constantly accelerating towards the center of the Earth, but since the velocity is always perpendicular to the acceleration, the net acceleration is zero and the satellite maintains a constant speed.

The kinetic energy of the satellite can be calculated as:

K = \frac{1}{2}mv^2

Where v is the velocity of the satellite. In circular motion, the velocity can be related to the radius of the orbit and the gravitational constant as:

v = \sqrt{\frac{GM}{R+h}}

Combining these equations, we get the total energy of the satellite as:

E = U + K = -\frac{GMm}{R+h} + \frac{1}{2}mv^2

Substituting the value of v, we get:

E = -\frac{GMm}{R+h} + \frac{1}{2}m\left(\sqrt{\frac{GM}{R+h}}\right)^2

Simplifying this equation, we get:

E = -\frac{GMm}{2(R+h)}

This is the total energy of the satellite in circular orbit at a distance R+h from the center of the Earth. Now, if we want to launch the satellite from the surface of the Earth (at a distance of R from the center), we need to provide it with enough energy to overcome the Earth's gravitational pull and reach the desired orbit.

This additional