Total energy of elliptical orbit

In summary, a comet is in an elliptical orbit around the sun. Its closest approach to the sun is a distance of 5e10 m (inside the orbit of mercury), at which point its speed is 9e4 m/s. Its farthest distance from the sun is far beyond the orbit of Pluto. Its speed when it is 6e12 m from the sun is 9e4 m/s.
  • #1
E92M3
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Why is the total energy of an elliptical orbit given by:
[tex]E_{tot}=\frac{-GMm}{2a}[/tex]
Where a=semi major axis.
I agree for a circular orbit I can do the following:
[tex]F_c=F_g[/tex]
[tex]ma_c=\frac{GMm}{r^2}[/tex]
[tex]\frac{v^2}{r}=\frac{GM}{r^2}[/tex]
[tex]v^2=\frac{GM}{r}[/tex]
Since the total energy also equal to the kinetic plus potential energy we have:
[tex]E_{tot}=\frac{1}{2}mv^2-\frac{GMm}{r}=\frac{1}{2}m\frac{GM}{r}-\frac{GMm}{r}=\frac{-GMm}{2r}[/tex]
Ok this is a similar form for circular orbit. But how can we just put a in instead of r for elliptical orbit? What is the justification?
 
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  • #2
Consider the fact that the total energy of the orbiting object is constant for any point of its orbit. Then take the visa-vis equation for finding the orbital speed at any point of an elliptical orbit:

[tex]\sqrt{GM \left (\frac{2}{r}-\frac{1}{a} \right )}[/tex]

Note that when r=a, you get

[tex] v=\sqrt{\frac{GM}{a}} [/tex]

or

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

Thus the total energy of a elliptical orbit with a semi-major axis of 'a' is the same as a that for a circular orbit with a radius of 'a'.
 
  • #3
Well actually I was trying to derive the vis-visa equation through the total energy. So I must have an alternate way to explain it or i'll be circular logic.
 
  • #6
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 5e10 m (inside the orbit of Mercury), at which point its speed is 9e4 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6e12 m from the Sun?

please show the equations needed, if you feel generous work the problem out as well.
 

FAQ: Total energy of elliptical orbit

1. What is the total energy of an elliptical orbit?

The total energy of an elliptical orbit is the sum of the kinetic energy and potential energy of the orbiting object. It is a combination of the object's speed and position in the elliptical orbit.

2. How is the total energy of an elliptical orbit calculated?

The total energy of an elliptical orbit can be calculated using the equation E = (1/2)mv^2 - GMm/r, where E is the total energy, m is the mass of the orbiting object, v is the velocity, G is the gravitational constant, M is the mass of the central body, and r is the distance between the two objects.

3. What factors can affect the total energy of an elliptical orbit?

The total energy of an elliptical orbit can be affected by the mass of the orbiting object, the mass of the central body, the distance between the two objects, and the speed of the orbiting object. Any changes in these factors can alter the total energy of the orbit.

4. How does the total energy of an elliptical orbit compare to a circular orbit?

The total energy of an elliptical orbit is usually greater than that of a circular orbit, as the elliptical orbiting object has both kinetic and potential energy, while a circular orbit only has kinetic energy. The total energy of a circular orbit is constant, while the total energy of an elliptical orbit varies as the object moves through different positions in the orbit.

5. Can the total energy of an elliptical orbit be negative?

Yes, the total energy of an elliptical orbit can be negative. This indicates that the orbiting object has a bound orbit, meaning it will continue to orbit the central body rather than escaping into space. A positive total energy would indicate an unbound orbit, where the object would eventually escape the orbit and move away from the central body.

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