# Satellite Orbiting Earth

## Homework Statement

A satellite of mass m kg is sent into orbit at a distance of 5R, where R is the radius of the earth. Write down an expression for:

a) The potential energy of the satellite in orbit.
b) The kinetic energy of the satellite in orbit.
c) The minimum work needed to sent the satellite from rest on earth in orbit.

## Homework Equations

$$GPE = \frac{MmG}{R^2}$$
M = Mass of Earth
$$Kinetic Energy = \frac{1}{2}mv^2$$
$$Work = F*D$$

## The Attempt at a Solution

a) Rewrite the formula for GPE.
b)Kinetic = -1/2Potential, rewrite formula for Gpe adding a division by -2.
c)$$W = FD$$
$$W = F5R$$

The force is the sum of the force needed to reach escape velocity on earth and that needed to keep the satellite in orbit.

The potential energy at rest of earth's surface is given by $$-\frac{MmG}{R}$$

As the kinetic energy has to be equal to reach escape velocity: $$\frac{1}{2}mv^2 = -\frac{MmG}{R}$$

How do I calculate the energy needed to keep it in orbit?

## Answers and Replies

Delphi51
Homework Helper
It will NOT reach escape velocity. If it did, it would not be in orbit.
You can't have negative kinetic energy, so the last formula you wrote cannot be correct.

Likely it is assumed that the orbit is circular, so you could actually calculate the velocity. Begin with
Centripetal Force = Force of Gravity
Fill in the detailed equations and solve for v.
Don't put the actual numbers in too early because you will want to compare with the potential energy -GMm/R

Basically, once it is orbit, I need to have potential energy equal to kinetic energy. Therefore:

$$\frac{1}{2}mv^2 = \frac{GMm}{R}$$
Where R equals the distance from the center of the earth.
And from there deduce an equation for v.

On the earth I have GPE equal to:

$$-\frac{GMm}{r}$$

What value should kinetic energy have to bring the satellite to a certain height? If I set it up such that PE = KE it will reach escape velocity and won't stop in orbit but, rather, continue (theoretically) to infinity.

Also, could someone explain me:

$$\frac{GMm}{R^2}$$

If I have understood correctly it is the force of attraction between two bodies, how does it differ from $$\frac{GMm}{R}$$?

Delphi51
Homework Helper
I wish I had your skill with LaTex! It doesn't seem to click with me.

The GMm/R² is the force of gravity on M and m - in this case, the centripetal force holding the satellite in orbit. If Work = integral of F*dr is integrated from R to infinity, you get -GMm/R which is the work required to lift the mass from radius R to infinite radius. It is called the potential energy of the mass due to its position near the Earth. Note that PE = 0 when radius is infinity, and PE is negative at all other radii.

The basic idea of a satellite in circular motion is that the gravitational force provides the centripetal force:
GMm/R² = mv²/R
If you cancel the R's and multiply by 1/2 you get
GMm/R = ½mv²
which is the same as the formula you have in your last post - note that it says the KE equals -PE rather than KE = PE. From its origin, you can tell that this condition has nothing to do with escape velocity; rather it is the condition for a stable circular orbit at radius R.
You could use it to answer part (b).

For part (c) you could begin with
E needed = Energy in orbit at Radius R - Energy at the surface
You already have the PE and KE for the orbit. At the surface, there is PE because of the definition - zero at infinite distance, negative energy at all other radii. You also have some KE because of the rotation of the Earth. Often this is ignored; you need to know the latitude of the launch site to calculate it accurately.

Hi, don't overestimate me, my skill with latex goes little beyond fractions and easy stuff.

So, if I understood correctly, the force of attraction between masses M and m where the distance between the two masses is R is defined by:

$$\frac{GMm}{R^2}$$

However, as my Kinetic Energy has to Equal my potential energy, and I can express kinetic energy in the form of centripetal acceleration:

$$\frac{GMm}{R^2} = \frac{mv^2}{R}$$

Or, simplified:

$$\frac{GMm}{R} = mv^2$$

However, I get confused where you ask to multiply by one half:

If you cancel the R's and multiply by 1/2 you get
GMm/R = ½mv²

Shouldn't I obtain $$\frac{GMm}{2R} = \frac{1}{2}mv^2$$

?

Hence summarizing:

PE = $$\frac{GM}{R^2}$$
KE = $$\frac{mv^2}{R}$$ or $$\frac{1}{2}mv^2$$ (The latter supposing to having cancelled R2 into R in PE.)

However, I'm still confused about the energy needed to bring it to height R. The PE at surface on earth is $$\frac{GM}{R^2}$$ where R is the radius of earth, a KE equal to PE would be escape velocity, should I simply subtract PE in orbit from PE on surface and state that an equal but opposite KE is needed?

Thanks in advance! :D

Delphi51
Homework Helper
Yes, ½mv² = GMm/(2R)
Sorry I lost the 2 on the right!
So the KE is not equal to the PE. In fact the KE equals -½ times the PE when in a circular orbit.

energy needed = energy up there - energy on earth
= (½mv² - GMm/R) - (-GMm/r)
where little r is the radius of the Earth. You can save yourself some calculating if you replace ½mv² with GMm/(2R).

energy needed = energy up there - energy on earth
= (½mv² - GMm/R) - (-GMm/r)

Wait, this way I'm subtracting KE in orbit from PE in orbit and then again subtracting PE on earth. Shouldn't it be PE(Orbit)-PE(Earth) = KE(Earth to Orbit)?

Delphi51
Homework Helper
It has both PE and KE when in orbit, totaling (½mv² - GMm/R).
That isn't subtracting; it is adding the negative PE.

So what I basically want is the difference between the combined PE and KE in orbit and the PE on earth, right? That will be the (negative) value my KE needs to have to reach orbit? But why do I want to include Orbital KE in my calculations? I am not interested only in orbital PE and Surface PE?

Delphi51
Homework Helper
KE is always positive.
If you only want to lift the satellite up to radius R with out any orbital velocity, then you need only find the PE. If you want it in orbit, you must add the KE necessary to give it the proper speed.

Basically, if r is the radius of earth and R is the distance from the center of the earth at a given point, the energy needed to lift a mass m up to a distance R is given by:

(Potential Energy at R) - (Potential Energy at r), expressed in terms of kinetic energy, hence:

$$(\frac{-GMm}{R^2}-\frac{-GMm}{r^2})$$

Therefore:

$$\frac{1}{2}mv^2 = (\frac{-GMm}{R^2}-\frac{-GMm}{r^2})$$

While the energy required to keep the satellite in orbit in given by a kinetic energy sufficient to balance the gravitational potential due to earth's attraction, correct?

Therefore the energy to keep it in orbit is:

$$\frac{-GMm}{R^2} = \frac{mv^2}{R}$$

I then have to sum up the two kinetic energies, correct?

Delphi51
Homework Helper
In the first part you wrote that ½mv² = PE at R - PE at r.
This is not correct because you have not included the KE when in orbit at R. The question clearly says "put into orbit" so you need the KE.

The last formula you wrote is wrong because the left side is negative and the right side is positive. If you drop the minus sign, it says the force of gravity on the satellite equals its centripetal force.

In the first part of my answer I was considering only the energy needed to bring it to orbital height disregarding the fact that KE is needed for it to remain up there. If that was the case, (Eg: Satellite can happily fall back on earth.), would Energy = Potential(Orbit) - Potential(Earth) be correct?

The second part is concerned with keeping it in orbit, given that it is already at orbital height I need the kinetic energy to be equivalent to the potential energy, right? So as I already have my PE I simply need an equivalent centripetal acceleration.

One question, for the second part I already know how much energy I need (That is, equivalent to PE), if it asked me to find only energy, would I still need to used centripetal acceleration? Aren't the formulas $$\frac{1}{2}mv^2$$ and $$\frac{mv^2}{R}$$ used to find velocity?

Delphi51
Homework Helper
Okay to the first paragraph.
I need the kinetic energy to be equivalent to the potential energy, right?
We keep going around on that. Better you look at another source. On this link look for the quoted sentence at the top of the last screenfull on the page and its derivation:
that is, the Kinetic Energy = -1/2 (Potential Energy) so the total energy in a circular orbit is half the potential energy.
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravPotEnergy.htm

Yes on your 3rd paragraph. I used that approach to derive the relationship between the kinetic and potential energies for an orbit - the same ones in the quote above.

You will also be interested in this earlier thread about the same question you have:
https://www.physicsforums.com/showthread.php?t=209261