Solve Orbital Mechanics Problems: Calculate Satellite Energy

In summary, the potential energy of an orbiting satellite is different from the mgh that applies only to something located close to the surface of a planet. Gravity decreases with distance, so the potential energy must be multiplied by the mass of the orbiting object to get the total potential energy.
  • #1
Praestrigiator
27
2
I'm having trouble with orbItal mechanics, I'm trying to determine the total potential energy of an orbiting satellite, what I've done so far is this:
I know m*g*h is potential energy, but I also know that gravity deceases with distance avoiding to the inverse square law. I know I Just can't use the satellites current Ag because that will increase as it falls towards the earth, and I'm having trouble determining exactly what that is.
This is giving me serious trouble because I want to calculate periapsis or apoapsis given just one of those two, and the velocity at that point.
 
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  • #2
Praestrigiator said:
I'm having trouble with orbItal mechanics, I'm trying to determine the total potential energy of an orbiting satellite, what I've done so far is this:
I know m*g*h is potential energy, but I also know that gravity deceases with distance avoiding to the inverse square law. I know I Just can't use the satellites current Ag because that will increase as it falls towards the earth, and I'm having trouble determining exactly what that is.
This is giving me serious trouble because I want to calculate periapsis or apoapsis given just one of those two, and the velocity at that point.
The gravitational potential energy of an orbiting body is different from mgh, which applies only to something located close to the surface of a planet, like earth.

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

U = -GMm / r

where

M - Mass of the planet, kg
m - mass of the orbiting body, kg
G - Universal Gravitational Constant = 6.67408 × 10-11 m3/kg-s2
r - distance between the centers of mass of M and m, in meters
U - gravitational potential energy (= 0 when r → ∞)
 
  • #3
I'd make one adjustment to SteamKing's equation. Use specific energy per unit of mass.

F= -GMm/r^2
F=ma
therefore, a=GM/r^2

All objects accelerate at the same rate (due to gravity).

Since it's the motion you're interested in, you can do the same with the energy (divide out the mass). And, if you do need to know the energy (to determine how much fuel to do a delta V, for example), you just multiply the specific energy (or the change in specific energy) by the mass when you need that info.

Likewise, your specific kinetic energy would be (v^2)/2. And your total specific energy would be (GM)/(-2a), where a is the semi-major axis of your orbit.

Given that you mentioned Earth, I assume you're talking about a satellite orbiting Earth? The universal gravitational constant and the mass of the Earth are constant. Once you've multiplied them together once, you can just remember the answer. You can even look up the answer in a book. It's your geocentric gravitational constant (3.986 x 10^5 km^3/sec^2).
 
  • #4
Is it really that simple? It makes sense but It seems too simple. I came up with that about two weeks ago but I didn't believe myself when I did it. I guess I'll just have to launch a scientific mission in KSP to verify it.
 
  • #5
Wait, that doesn't make sense, by getting higher your potential energy decreases? And in a higher orbit you move slower so your kinetic energy decreases too, that can't be right.
I'm looking for Σ specific potential energy, our the whole of all the specific kinetic energy that would be gained from a fall of such altitude.
 
  • #6
Praestrigiator said:
Wait, that doesn't make sense, by getting higher your potential energy decreases? And in a higher orbit you move slower so your kinetic energy decreases too, that can't be right.
I'm looking for Σ specific potential energy, our the whole of all the specific kinetic energy that would be gained from a fall of such altitude.

First, you were looking for orbital elements; now, you're looking for the K.E. a satellite gains by falling out of orbit.

This article discusses orbital mechanics in some detail:

http://www.braeunig.us/space/orbmech.htm

This article talks about the relationship between gravitational P.E. into K.E. for an orbiting body:

https://en.wikipedia.org/wiki/Specific_orbital_energy
 
  • #7
Praestrigiator said:
Wait, that doesn't make sense, by getting higher your potential energy decreases? And in a higher orbit you move slower so your kinetic energy decreases too, that can't be right.
I'm looking for Σ specific potential energy, our the whole of all the specific kinetic energy that would be gained from a fall of such altitude.
No. Notice the minus sign in front of the expression fro PE.
The PE increase as you go higher, the maximum value being zero. This is so because the reference is chosen to be at infinite distance.
 
  • #8
SteamKing said:
First, you were looking for orbital elements; now, you're looking for the K.E. a satellite gains by falling out of orbit.

This article discusses orbital mechanics in some detail:

http://www.braeunig.us/space/orbmech.htm

This article talks about the relationship between gravitational P.E. into K.E. for an orbiting body:

https://en.wikipedia.org/wiki/Specific_orbital_energy
I don't see how my request has changed. if an object has 25,000,000 Joules of kinetic energy when it impacts, then it must have had 25,000,000 Joules of potential energy when it began to fall. I am just looking for the potential energy of an object in freefall beginning at a distance where the gravitational inverse-square law becomes relevant, as I always have been. maybe I was sloppy with my wording before, but I truly do not see a difference between what I was originally asking for and the description you gave here.
 
  • #9
Praestrigiator said:
I don't see how my request has changed. if an object has 25,000,000 Joules of kinetic energy when it impacts, then it must have had 25,000,000 Joules of potential energy when it began to fall. I am just looking for the potential energy of an object in freefall beginning at a distance where the gravitational inverse-square law becomes relevant, as I always have been. maybe I was sloppy with my wording before, but I truly do not see a difference between what I was originally asking for and the description you gave here.
The gravitational inverse square law is always relevant, regardless of distance. After all, this law still works between Pluto and the Sun, even though these two bodies are separated by a distance of some 4.5 billion kilometers.

If you want to calculate the periapsis or apoapsis of an orbiting body, then application of Newton's laws of motion can tell you this.

If you want to calculate the impact energy of a body falling out of orbit, a different application of the same laws is required.

Depending on the problem you want to solve, you have to tailor the analysis to obtain the solution. It is not immediately obvious how knowing the periapsis of an orbiting body will necessarily convert into the K.E. of that body hitting the ground.
 
  • #10
Praestrigiator said:
I don't see how my request has changed. if an object has 25,000,000 Joules of kinetic energy when it impacts, then it must have had 25,000,000 Joules of potential energy when it began to fall.
No, it must have had 25,000,000 more Joules of potential energy when it began to fall than when it impacts. Zero is 25,000,000 more than -25,000,000.
 
  • #11
Praestrigiator said:
I'm having trouble with orbItal mechanics, I'm trying to determine the total potential energy of an orbiting satellite, what I've done so far is this:
I know m*g*h is potential energy, but I also know that gravity deceases with distance avoiding to the inverse square law. I know I Just can't use the satellites current Ag because that will increase as it falls towards the earth, and I'm having trouble determining exactly what that is.
This is giving me serious trouble because I want to calculate periapsis or apoapsis given just one of those two, and the velocity at that point.

The total orbital energy of your satellite is the sum of it's kinetic and potential energies.
KE = mv2/2
and
GPE = - GMm/r
(note since r is measured from the center of M, working out the potential energy difference between the satellite in orbit vs on the ground means solving the following
[tex]\Delta GPE = \frac{GMm}{r_E}- \frac{GMm}{r_o}[/tex]

Where rE and ro are the Earth's radius and orbit radial distance

Now, to get to your orbital question:

We start with:

(1)[tex]E= \frac{mv^2}{2}- \frac{GMm}{r}[/tex]

If we assume a circular orbit, we know that [itex]v= sqrt{\frac{GM}{r}}[/itex]

subbing for v, and reducing the equation gives

(2)[tex]E=- \frac{GMm}{2r}[/tex]

It also turns out that this equation holds for elliptical orbits if we use a, the semi-major axis for r:

(3)[tex]E=- \frac{GMm}{2a}[/tex]

If we equate equation 1 and 3 and solve for a, we get

[tex]a =\frac{1}{\frac{2}{r}-\frac{v^2}{GM}}[/tex]

which gives us the semi-major axis if we know the velocity at r
(This last equation is a re-arrangement of the vis viva equation)

Now its just a simple case of knowing that

[tex]a= \frac{r_{per}+r_{app}}{2}[/tex]

to find either apogee or perigee if you know the other and the velocity at it.
 

1. What is orbital mechanics and why is it important?

Orbital mechanics is the study of the motion and behavior of objects that are in orbit around a larger body, such as a planet or star. It is important because it helps us understand and predict the movement of satellites, spacecraft, and other objects in space, which is crucial for successful space missions and satellite operations.

2. How do you calculate satellite energy?

Satellite energy is calculated using the formula: E = -G(m1m2)/2r, where G is the universal gravitational constant, m1 and m2 are the masses of the two objects (in this case, the satellite and the body it is orbiting), and r is the distance between the two objects. This calculation takes into account the gravitational potential energy of the satellite in its orbit.

3. What factors affect a satellite's energy?

The main factors that affect a satellite's energy are its mass, the mass of the body it is orbiting, and the distance between the two objects. Other factors that can influence a satellite's energy include atmospheric drag, solar radiation, and the gravitational pull of other nearby objects.

4. Can you explain the difference between kinetic and potential energy in orbital mechanics?

Kinetic energy in orbital mechanics refers to the energy that an object in orbit has due to its motion. This can be calculated using the formula: KE = (1/2)mv^2, where m is the mass of the object and v is its velocity. Potential energy, on the other hand, is the energy that an object has due to its position or location in relation to other objects. In orbital mechanics, potential energy is mainly related to the gravitational pull of the larger body it is orbiting.

5. How is satellite energy used in real-world applications?

Satellite energy is used in a variety of real-world applications, including telecommunications, weather forecasting, navigation, and scientific research. By understanding and calculating satellite energy, we can accurately predict the orbit and behavior of satellites, ensuring their proper functioning and longevity in space.

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