# Direction of Burn while in Orbit

1. May 3, 2015

### PPERERA

I know that when a spacecraft is in orbit, the direction in which the spacecraft produces thrust affects its orbit.

If the thrust is tangent to the orbit, the orbit's shape (eccentricity and size) changes. If the thrust is perpendicular to the orbit, but tangent to the surface of the object being orbited, the orbit's inclination changes.

But what happens if the thrust is directed radially from the surface of the object being orbited?

2. May 3, 2015

### rcgldr

As you mentioned, if the thrust is perpendicular to the path, then no work is done, the total energy remains the same and just the path changes. If the thrust is radial, but the path is not perpendicular to a radial line, then some work is done, changing the total energy. You could split up the radial thrust into components in the direction of (parallel) and perpendicular to the path.

3. May 3, 2015

### Staff: Mentor

If you do that in a circular orbit, you will also change eccentricity, but firing in the direction of motion (at a different position to get the same argument of periapsis) is much more effective.

4. May 12, 2015

### PPERERA

[Mentor's Note: Post merged from another thread.]

In a previous question (https://www.physicsforums.com/threads/direction-of-burn-while-in-orbit.811953/), I learned that burning the engine radially relative to the object being orbited had the same effect as burning parallel to the tangent of the orbiting craft's velocity. In both cases, the shape of the orbit changes.

In addition to this, I know that burning the engine parallel to the orbital motion changes the eccentricity of the orbit. Is the same true when the engine is burned radially?

Imagine a spacecraft in a circular orbit. It burns its engine radially away from the the planet. Does its orbit remain a circular orbit while expanding its radius, or does it become elliptical?

An extension to this question:
If it is true that burning radially does not change the shape eccentricity of the orbit, would it be useful to burn radially to change orbits rather than burning parallel to the orbital motion? When I ask this, I'm considering not just the amount of propellant that has to be carried up, but also the load that conventional orbital transfers put on the computing systems and the mission control and the reliability of the both types of transfers (radial and parallel).

Last edited by a moderator: May 14, 2015
5. May 12, 2015

### PPERERA

Thanks for the answer. But wouldn't producing thrust radially and perpendicular to the orbit still do work against the gravity of the object being orbited?

6. May 13, 2015

### rcgldr

Start off with a circular orbit. If an outwards perpendicular force is applied to the object, it's path becomes non-circular, and once the path is non circular (and outwards) a component of gravity will slow the object down. If the force remains perpendicular to the object's motion, then no work is done, so this only changes the shape of the orbit. The distance can no longer be increased once the objects speed is reduced to zero, which happens when the objects gravitational potential energy equals it's total energy.

Take the case where the perpendicular force is continuously adjusted so that the object travels in a straight line that doesn't intersect the earth. The force is perpendicular to the objects motion, so the force does not affect the objects speed. However as the object increases distance from the earth, the direction of gravity opposes the objects motion, but with decreasing strength (G M m / r^2). Assuming the object is not initially traveling at or above escape velocity, it slows down and eventually stops, and then begins a return path. The perpendicular force could be adjusted to continue to maintain a straight line, while the object moves back and forth along the straight line.

If the outwards force is radial, then once the object's path is no longer circular, the outwards radial force opposes gravity, so the object slows down less or if the force is great enough, the object speeds up. If you think of this as a single impulse to produce some amount of radial velocity, then the final velocity = sqrt(original velocity^2 + radial velocity^2). If the the force was in the direction of the object, than the final velocity = original velocity + change in velocity, which is more efficient. Say you wanted to double the velocity, with the radial thrust, the radial velocity would need to be sqrt(3) ~= 1.73 x original velocity, while with thrust in the direction of motion, the change in velocity would need to be 1.00 x original velocity.

Last edited: May 13, 2015
7. May 13, 2015

### BobG

While you're thrusting, you'll have acceleration in a radial direction. Multiply that acceleration by the amount of time you were thrusting, and you'll get a change in velocity. Add that change in velocity to the original velocity (keeping in mind the two components are perpendicular to each other) and you'll get your new velocity vector.

Your new orbit will be tangent to that new velocity vector.

Now lots of stuff about your orbit have changed. You rotated your orbit, which moves your perigee points and apogee points. You've also changed the shape of your orbit and the size of your orbit. Additionally, since your perigee point moved, but your spacecraft has barely moved at all (yet), your position relative to perigee (true anomaly) has changed.

Other than that, it's the same as thrusting in the same direction the spacecraft is moving. Except if your change in velocity is the same direction as your original velocity, less things about your orbit change since your change in velocity didn't rotate your orbit (provided your in-track change occurred at perigee or apogee when the velocity vector was perpendicular to the radius - there are no other points where an in-track change wouldn't have at least some radial component to the delta-v).

Last edited: May 13, 2015
8. May 13, 2015

### rcgldr

Except there is less of a net increase in velocity. For the radial case, the final velocity = sqrt((original velocity)^2 + (change in radial velocity)^2). For the same direction case, the final velocity = original velocity + change in velocity.

9. May 13, 2015

### Staff: Mentor

[Mentor's Note: Post merged from another thread]

Imagine a short burn - the spacecraft won't move much during the burn. Can it be on a circular orbit significantly further outside afterwards? Of course not.
It is not true, and even if it were, it would not be useful. Thrusting in the direction of motion or against is more effective unless you have to change the orbital plane (because the inclination is wrong, for example).

Last edited by a moderator: May 14, 2015
10. May 14, 2015

### BobG

It's generally not true that a radial burn would not be useful, but that's assuming fuel is your primary concern. If time is your primary concern, a fast transfer requiring a radial burn would be effective, just at a large cost in fuel.

In almost all cases, fuel is your primary concern since you actually have to launch all the fuel you need into orbit along with your spacecraft. (So I'm nitpicking a bit, but just in the interest of thoroughness.)

11. May 14, 2015

### QuantumPion

A radial burn will increase the altitude of the point of the orbit 90° away and decrease the altitude of the point of the orbit 270° away (or vice-versa for anti-radial burn). Radial burns can be useful for adjusting the timing of orbit to rendezvous with another craft or planet. For example, if you want to Hohmann transfer to a target, but are too far ahead or behind, you can make a radial burn to advance or delay the time you arrive at the intercept point.