# Orbital Motion Help

1. Nov 16, 2004

### chilli_pepper

Hi guys and gals

I have been given this question and have spent so far around 4 hours pondering calculating researching and reading and have got no where except into a large state of confusion!

Its probably really really simple but i just cant seem to get logical answers and now i have done so many methods i dont know where i am.

Do you mind giving me some guidance? The question is below:

A spacecraft of mass m = 80000 kg is initially in a circular meridianal orbit of altitude H0 = 300 km above the earth’s surface. When the spacecraft is in the position A above the North Pole its two orbital-manoeuvring-system (OMs) engines, each of which has a trust of F = 25 kN, are fired during the time period dt = 35.5 seconds to increase the velocity of the spacecraft and thus to transfer it to the new elliptical orbit. Calculate the altitude H1 of the spacecraft at the apogee point B of the new elliptical orbit. Ignore the change in position of the spacecraft while the engines are on.

Use the following values for the other required parameters:
Radius of the earth: R = 6371 km
Gravity acceleration: g = 9.825 m/s2

the nearest i have is ~410km but got that using hohmann transfer and data that is not given in this question so doesnt really count!

Any help greatly appreciated

Hope this is the right forum to post it :S its my first post! :)

Dan

2. Nov 17, 2004

### tony873004

What did you get for your spaceship's delta V?

Why is using Hohmann transfer formulas cheating?

Are you forbidden from using any data not given, such as the Gravitational Constant or the Earth's mass?

Hohmann formulas use mu which is G*M, and without knowing either G or M, you can still compute mu with the info given.

a = mu (1/d^2)
9.825 = mu (1 / 6371^2)
mu = 398793222.825

I got a different final answer than you did. Post your formulas. I may be wrong. :yuck:

Last edited: Nov 17, 2004
3. Nov 17, 2004

### Jenab

I'll post his formulas.

Re = 6371000 meters

R = Re + 300000 meters
R = 6671000 meters

GM = 3.98793E+14 m^3 sec^-2

Vo = sqrt(GM/R)
Vo = 7731.8 m/s

F = 2 (25000 kg m sec^-2)
m = 80000 kg

acc = F/m
acc = 0.625 m sec^-2

t = 35.5 sec

dV = (acc) (t)
dV = 22.1875 m/s

Vf = Vo + dV
Vf = 7753.9 m/s

a = 1 / { 2/R - (Vf)^2 / GM }
a = 6709564 meters

The perigee distance is the same as the burn distance:

Rp = R
Rp = 6671000 meters

e = 1 - Rp/a
e = 0.005748

Ra = a (1+e)
Ra = 6748128 meters

apogee altitude = Ra - Re
apogee altitude = 377.1 kilometers

Jerry Abbott

4. Nov 17, 2004

### Jenab

Or, to put it into one long equation...

apogee altitude = 2 / { 2/Rp - [sqrt(GM/Rp) + tF/m]^2 / GM } - Rp - Re

where

Rp is the geocentric distance of the spaceship at perigee (meters)
Re is the Earth's radius (meters)
t is the duration of burn (seconds)
F is the total thrust during burn (Newtons)
m is the mass of the rocket during burn (kilograms, assumed constant)
GM is the gravitational parameter of Earth (m^3 sec^-2)

apogee altitude comes out in meters

Rp is equal to the radius of the original circular orbit

transit of the spaceship during burn is ignored

The billygoat goes back to the barn

Jerry Abbott

Last edited: Nov 18, 2004
5. Nov 17, 2004

### tony873004

I get the same answer you do, 377 km altitude, using a simulator instead of formulas.