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rieman zeta

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- Thread starter rieman zeta
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- #1

rieman zeta

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- #2

marlon

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1) conservation of total energy

2) conservation of angular momentum

Check out : http://www.dur.ac.uk/bob.johnson/SL/

regards

marlon

- #3

rieman zeta

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could you be a little less opaque please. jbf

- #4

marlon

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What i wanted to say with my first post is the fact that all the formula's that you use will be a result of applying the two mentioned laws onto the orbits and object's motion.

marlon

- #5

rieman zeta

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sir: A spacecraft with rocket capability approaches the sun from infinity in a parabollic orbit. It has a twin folowing. The first fires its rockets tangent to its trajectory and opposite the direction of motion at 3 times the radius of the Sun. Later, the second waits until it is at perihelion (2 solar radii) before firiing an identical burn, force x del t. Why is it when they compare their velocity excess at infinity is the second's greater than the first?

This is a strictly two body problem done in heliocentric coordinates.

Of course it is your choice to give an answer couched in whatever orbital mechanics equations seem to bear. Ususally there is a direct explanation related to a well recognized physical principle.

If there is some necessity to frame the question with more clarity please advise again.

jbf

- #6

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Assume that the rocket burns very fast, so it essentially gives an impulse to velocity.

As another poster stated, there are two important parameters that determine the shape of the orbit, angular momentum and energy.

In this case, we are concerned with the total velocity at infinity, which is initially zero.

"Velocity at infinity" is just another name for energy, so we are concerned with the rocket's total energy, which is assumed to be conserved - this will be exactly true in center-of-mass, i.e. heliocentric, coordinates if the rocket doesn't interact with any other planets, and approximately true if it doesn't have any close encounters with other planets.

The rocket gains the most energy per unit velocity change when it thrusts in its direction of motion (which has been already assumed), and when it is moving the fastest.

This is because energy is not linear, it's .5*m*v^2 + V(r), where V(r) is the potential energy, and v is the rocket's velocity.

To illustrate: a rocket at a velocity of 6 has 18*m units of energy, at a velocity of 8 it has 32*m units of enregy, and at a velocity of 10 it has 50*m units of energy. The rocket which accelerates from a velocity of 6 to 8 gains 14*m units of energy, while the rocket that accelerates from a velocity of 8 to 10 gains 18*m units of energy.

- #7

BobG

Science Advisor

Homework Helper

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You get a better picture if you start with an elliptical orbit. That way you can compare apogee, perigee, and some point between.

What you'll find is that the change in angular momentum is directly proportional to the change in velocity.

If perigee is 7500km and the initial velocity is 8928.6 m/s, then the angular momentum is 66964.6 km^2/s. If you increase the velocity by 1000 m/s, the satellite is going 1.112 times as fast as its initial perigee velocity. The angular momentum is also 1.112 times as great, or 74464.6 km^2/s. The eccentricity**increases** from .5 to .8548 and **apogee** of the orbit increases from 22500 km to 95816 km. Perigee remains unchanged since that's where the maneuver occurred.

If apogee of the same orbit is 22500 and the apogee velocity is 2976.2 m/s, then the angular momentum is still 66964.6 km^2/s (angular momentum is conserved). If you increase the velocity by 1000 m/s, the satellite is going 1.336 times as fast and the angular momentum is 1.336 times greater, or 89464.6 km^2/s. The eccentricity of the orbit**decreases** from .5 to .10755 and **perigee** increases from 7500 to 18130 km. Apogee remains unchanged since that's where the maneuver occurred.

You can probably anticipate what's going to happen if you do the maneuver at the covertex (true anomaly equals 120 degrees). Before the maneuver, your**speed** (now there's two components to the velocity since you're not at perigee or apogee) is 5154.9 m/s with angular momentum conserved. Increasing the speed by 1000 m/s results in a speed 1.194 times as fast and angular momentum is 1.194 times as great, or 79954.96. The eccentricity of the orbit **increases**, but only to .62117. **Both** perigee and apogee increase - perigee to 9892.9 km and apogee to 42336.4 km.

A maneuver at the opposite covertex (as you go from apogee to perigee) gives the exact same results with one difference. At 120 degrees, the argument of perigee progresses and true anomaly regresses. At 240 degrees, the argument of perigee regresses while true anomaly progresses.

The further away from the Earth you perform a maneuver, the bigger effect it will have on the orbit. However, performing the maneuver at perigee focuses all of the effect towards the opposite side - hence a greater impact on apogee even though the effect on the orbit overall is less.

Edit: I reread your more detailed post. Once you've set your angular momentum, it remains constant. A maneuver at perigee has the least affect on the orbit's angular momentum, but increases apogee. That means there must be a proportional decrease in velocity at apogee if angular momentum will remain constant. An given increase in speed at perigee will result in a lower velocity at apogee than an increase anywhere else in the orbit.

Except your specific example talked about decreasing the speed instead of increasing it. If the speed is decreased, the effects are the opposite. A decrease in speed has the least effect on the orbit's angular momentum, but the greatest effect on the apogee radius. Since angular momentum remains constant once set, there has to be an increase in apogee speed to offset the decrease in apogee radius.

What you'll find is that the change in angular momentum is directly proportional to the change in velocity.

If perigee is 7500km and the initial velocity is 8928.6 m/s, then the angular momentum is 66964.6 km^2/s. If you increase the velocity by 1000 m/s, the satellite is going 1.112 times as fast as its initial perigee velocity. The angular momentum is also 1.112 times as great, or 74464.6 km^2/s. The eccentricity

If apogee of the same orbit is 22500 and the apogee velocity is 2976.2 m/s, then the angular momentum is still 66964.6 km^2/s (angular momentum is conserved). If you increase the velocity by 1000 m/s, the satellite is going 1.336 times as fast and the angular momentum is 1.336 times greater, or 89464.6 km^2/s. The eccentricity of the orbit

You can probably anticipate what's going to happen if you do the maneuver at the covertex (true anomaly equals 120 degrees). Before the maneuver, your

A maneuver at the opposite covertex (as you go from apogee to perigee) gives the exact same results with one difference. At 120 degrees, the argument of perigee progresses and true anomaly regresses. At 240 degrees, the argument of perigee regresses while true anomaly progresses.

The further away from the Earth you perform a maneuver, the bigger effect it will have on the orbit. However, performing the maneuver at perigee focuses all of the effect towards the opposite side - hence a greater impact on apogee even though the effect on the orbit overall is less.

Edit: I reread your more detailed post. Once you've set your angular momentum, it remains constant. A maneuver at perigee has the least affect on the orbit's angular momentum, but increases apogee. That means there must be a proportional decrease in velocity at apogee if angular momentum will remain constant. An given increase in speed at perigee will result in a lower velocity at apogee than an increase anywhere else in the orbit.

Except your specific example talked about decreasing the speed instead of increasing it. If the speed is decreased, the effects are the opposite. A decrease in speed has the least effect on the orbit's angular momentum, but the greatest effect on the apogee radius. Since angular momentum remains constant once set, there has to be an increase in apogee speed to offset the decrease in apogee radius.

Last edited:

- #8

kjknohw

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- 0

example: going 3km/s your energy is 9 units. add 1 km/s and energy is 16. You get 7.

going 4km/s and adding 1km/s makes you gain 9 kinetic energy units

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