# A satellite changes course

1. Nov 24, 2012

### Karol

1. The problem statement, all variables and given/known data
A satellite of mass m rotates in a circle of radius a round a star of mass M.
At point P it's velocity V0 is altered to be parallel to the x axis.
The trajectory becomes an ellipse. show that the axes of the ellipse are a and b.

2. Relevant equations
Potential energy:
$$U=-\frac{gMn}{R}$$
Total energy of a satellite: Etot=U+Ek
The ellipse definition: the trajectory of a rope of fixed length, attached on both it's ends.

3. The attempt at a solution
The solution is that the kinetic energy increases monotonically from point A to point B, and so there is only one point, which is P, at which: Etot=-Ek=|U/2|.
And, from geometry, the distance from the ellipse focus to P equals a, which i understand, so, the ellipse's axes must be a and b.
What i don't understand is why the condition Etot=-Ek=|U/2| must occur specifically at point P.
I understand that U changes linearly with the distance R, and that U is symmetric around point P: at point A it is less the same amount that it is higher at point B, but still i think that the condition Etot=-Ek=|U/2| may occur at other points on the ellipse

#### Attached Files:

• ###### Sattelite.jpg
File size:
19.2 KB
Views:
47
2. Nov 24, 2012

### Staff: Mentor

Since the orbit change occurs from an initially circular orbit, the radius must be of length a. The velocity is also specified to be Vo for both the circle and the ellipse. So you can write the expression for the total energy (specific energy $\xi$).

You should also know an expression for the specific energy of an orbit in terms of its major axis and the gravitational parameter μ. Equate, isolate KE.

3. Nov 24, 2012

### Karol

I don't know what specific energy is, i know what potential and kinetic energies are, and that their sum is the total energy of the satellite.

4. Nov 24, 2012

### Staff: Mentor

Total energy in the Kepler problem depends on the semi-major axis only (and the masses and G of course) - the energy is not modified, so the semi-major axis of the ellipse is "a" as well.
Specific energy is just the total energy divided by the mass of the satellite - as it does not change its mass, it does not matter which one you use.

5. Nov 24, 2012

### Staff: Mentor

Specific energy is just the energy terms without the mass. Thus, for example, the specific kinetic energy is V2/2. Think of it as being the energy per unit mass for the object in question. The total specific energy for an orbit is

$$\xi = \frac{V^2}{2} - \frac{\mu}{r}$$

6. Nov 24, 2012

### Karol

I know only that the total energy is Etot=-GMm/2a because i started and derived this equation from a circular trajectory, i know nothing about total energy in the kepler problem.
Can this equation be derived for an elliptic orbit of it's own, without connection to a circular movement? then i will understand it relates to the semi-major axis of an ellipse.
And, what is a semi-major axis of an ellipse? isn't it a in my drawing?
And how is b called?

Edit by Borek: please attach only png/gif/jpg images, not bmp.

#### Attached Files:

• ###### Untitled-1.png
File size:
742 bytes
Views:
33
7. Nov 24, 2012

### Staff: Mentor

A circular orbit is a "special case" of an elliptical orbit where the eccentricity happens to be zero

Your equation for Etot includes the mass of the satellite, m. If you drop the m you have the specific mechanical energy for the orbit: $\xi = -\frac{G M}{2 a}$. Now, when M >> m then $\mu ≈ GM$. So that $\xi = -\frac{\mu}{2a}$.

This relates the specific mechanical energy of the orbit to the size of the semi-major axis of the orbit, a, which is half the length of the long axis of the ellipse as in your drawing. The perpendicular axis (across the "width" of the ellipse) is known as the minor axis. Half of it is b, the semi-minor axis.

So now you have two equations for the specific mechanical energy. Equate them and plug use the given information for the point P.

8. Nov 25, 2012

### Staff: Mentor

It can.

Half the major axis (the longer axis in the ellipse), and identical to the distance between focal points and the points at the ends of the minor axis (P in your sketch). This geometric identity can be used to show that P is really on the minor axis of the ellipse.

9. Nov 25, 2012

### Karol

How can the equation Etot=-GMm/2a be derived for an ellipse, without connection to a circle, and how can it be shown that a is the semi-major axis?

10. Nov 25, 2012

### Staff: Mentor

It involves a consideration of the constants of the motion that come from the trajectory equation for a body in orbit (the differential equation describing the motion). For a derivation, I refer you to a text on orbital mechanics. In particular, I recommend Fundamentals of Astrodynamics by Bate, Mueller, and White. It's a rather inexpensive softcover that covers the topic quite well.

11. Nov 27, 2012

### Karol

Why, when the velocity's direction changes, the ellipse is in the shape, size and position as in my drawing?
Why is it's major axis a, the same as the circle's radius?
I was told that since the kinetic energy grows, monotonically, from A to B, there is one point at which Etot=-Ek=|U/2|, and this is P.
Why, necessarily, P? it could occur on other points on the ellipse.

12. Nov 27, 2012

### Staff: Mentor

The problem states that the change makes the velocity parallel to the x-axis at point P. The Sun is the focus which lies on the x-axis. By symmetry then, the point P must lie at an end of the minor axis -- nowhere else on the ellipse is the trajectory parallel to the major axis.

When you construct an ellipse using two pins and a thread, the pins being stuck into the two foci, the thread with length 2a tied to the pins, and a pencil run around the perimeter bounded by the taut thread, then the sum of the two radii defined by the pencil-to-pin thread sections is a constant 2a.

In the figure, r + r' = 2a. Now, when the pencil reaches the vertical mid line of the ellipse, r = r' so then r = a.

#### Attached Files:

• ###### Fig1.gif
File size:
2.9 KB
Views:
101
13. Nov 27, 2012

### Karol

The situation could have been like this, with a the radius of the circle.
In this case the velocity is parallel to the initial x axis also, and a isn't the major axis.

#### Attached Files:

• ###### Ellipse.jpg
File size:
10.2 KB
Views:
34
14. Nov 27, 2012

### Staff: Mentor

No, the original diagram for the problem indicates that the major axis of the intended ellipse lies along the x-axis.

Also, not just any ellipse will do -- it must have a total energy consistent with the kinetic and potential energy at the instant it is created (at the course change). A body on a circular orbit of radius a has an orbital velocity of $v_o = \sqrt{\mu/a}$. So its specific kinetic energy, being $v_o^2/2$, is $\mu/(2a)$. The specific PE is $-\mu/a$. That means the total specific energy is
$$\xi = \frac{\mu}{2a} - \frac{\mu}{a}$$
This simplifies to
$$\xi = -\frac{1}{2}\frac{\mu}{a}$$
in other words, (minus) half the PE.

Last edited: Nov 27, 2012
15. Nov 28, 2012

### Karol

leave out the initial drawing, it tries to prove that the situation is in that manner.
Maybe it is, it could be different.
The ellipse doesn't necessarily create itself according to the drawings.

16. Nov 28, 2012

### Staff: Mentor

No matter what set of axes you choose, the initial kinetic and potential energies are fixed by the conditions of the problem; distance is the radius of the circular orbit, a, and speed is that of the circular orbit speed, $\sqrt{\mu/a}$. This fixes the energy of the orbit:

$\xi = \frac{\mu}{2a} - \frac{\mu}{a} = -\frac{\mu}{2a}$

Now, for all orbits the specific mechanical energy is related to the semi-major axis a by:

$\xi = -\frac{\mu}{2a}$

By inspection we conclude that the a's are the same in both cases, namely semi-major axis of the ellipse.