Orbit Semimajor Axis and Eccentricity after increasing V

Click For Summary
SUMMARY

The discussion focuses on calculating the new semimajor axis and eccentricity of a space shuttle's orbit after a velocity increase of 400 m/s at apogee. The original semimajor axis was 7000 km with an eccentricity of 0.05. After the velocity increase, the new semimajor axis is determined to be 7809 km. The eccentricity can be calculated using the formula e=(sma-perigee)/sma, where the perigee is derived from the original parameters.

PREREQUISITES
  • Understanding of orbital mechanics, specifically semimajor axis and eccentricity
  • Familiarity with gravitational parameters, including Earth's gravitational constant (u)
  • Knowledge of velocity calculations in orbital dynamics
  • Ability to apply the vis-viva equation for orbital energy
NEXT STEPS
  • Learn how to calculate orbital parameters using the vis-viva equation
  • Research the effects of delta V on orbital trajectories
  • Study the relationship between eccentricity and orbital shapes
  • Explore advanced orbital mechanics simulations and tools
USEFUL FOR

Aerospace engineers, astrophysicists, and students studying orbital mechanics will benefit from this discussion, particularly those interested in the effects of velocity changes on orbital parameters.

ryank614
Messages
8
Reaction score
0
Question:

A space shuttle is in an orbit about the Earth. At its apogee, it uses thrusters and increases its velocity by 400 m/sec. What is the new orbit semimajor axis, eccentricity and how much will the next perigee altitude be increased?

Known:

Original semimajor axis: 7000 km -> a
Original eccentricity: 0.05 -> e
Earth's Radius: 6378 km
u= GxEarth's Mass=3.986x10[tex]^{5}[/tex]

What I have done so far:

I figured out the apogee and perigee of the orbit, as well as the velocity at the apogee before the firing of the thrusters.

i) apogee: a(1+e) = 7350 km
ii) perigee: a(1-e) = 6650 km
iii) velocity at apogee:

[tex]\sqrt{u*((2/r)-(1/a))}/[/tex] where r = apogee.

I got v=7.17 km/s

Now after the thrusters are fired, the new velocity is 7.57 km/s

Using [tex]\epsilon[/tex] = V[tex]^{2}[/tex][tex]/2[/tex] - u[tex]/r[/tex] where r is the current position, aka the apogee and plugging [tex]\epsilon[/tex] into

a = -u[tex]/2\epsilon[/tex]

I found the new semimajor axis to be 7809 km. But then here is the problem. How do I find out the new eccentricity? Thanks!
 
Astronomy news on Phys.org
There's 1 assumption you're making, and that's that your 400 m/s of delta V are in the prograde direction. You could fire your engines to accelerate away from the Earth instead, until your total velocity increased by 400 m/s. Or you could fire towards the Earth, or anywhere inbetween.

But I imagine they're expecting you to make this assumption. This means that after a delta V of 400 m/s in a prograde burn, your new velocity will momentarily be all tangental. This means that you're either still at apogee, or at your new perigee. Just compute circular velocity for that distance and see if you're above it (which means you're at perigee), or below it (which means you're at apogee). Your now know your apogee or perigee and your semi-major axis, so eccentricity should be easy to find: e=(sma-perigee)/perigee, or e=(apogee-sma)/sma
 
Last edited:
I should have said e=(sma-perigee)/sma
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 62 ·
3
Replies
62
Views
8K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K