Orbit Transfer to Lower Apoapsis of a Mars Orbit

  • Thread starter ColdFusion85
  • Start date
  • Tags
    Mars Orbit
In summary: The rest is fine.In summary, to lower the apoapsis of a Mars mapping orbit, only one burn at periapsis is needed. The initial orbit has a periapsis radius of 3800 km and an eccentricity of 0.5, with the final apoapsis radius to be half of the initial one. Using the General Velocity Equation with the values for the gravitational constant (4.293 E4 km^3/s^2), and the radius of Mars (3397 km), the initial velocity is calculated to be 4.43 km/s and the final velocity is 4.03 km/s. This results in a required delta-V of -0.4 km/s,
  • #1
ColdFusion85
142
0

Homework Statement


Consider an orbit transfer to lower the apoapsis of a Mars mapping orbit. The initial orbit has a periapsis radius of 3800 km and an eccentricity of 0.5. The final apoapsis radius is to be 1/2 of the initial one. Note that only one burn at periapsis is needed to perform this change and that the periapsis radius of the final orbit is the same as the initial orbit. Find the [tex]\Delta V[/tex] required.


Homework Equations


General Velocity Equation: V=[tex]\sqrt{{2\mu/R}-{\mu/a}}[/tex], where [tex]\mu[/tex]=GM, and a is the semi-major axis

Relation of radii to semi-major axis: [tex]r_p + r_a[/tex] = 2a, where [tex]r_p[/tex] is the radius at periapsis, and [tex]r_a[/tex] is the radius at apoapsis


The Attempt at a Solution


First I calculated the radius at initial apoapsis using the formula [tex]r_a = r_p((1+\epsilon)/(1-\epsilon))[/tex]

I got 11400 km, which means the final apoapsis is 5700 km.

Using the relation for radii to semi-major axis I calculated the initial semi-major axis and got 7600 km. Doing the same for the final semi-major axis, I got 4750 km.

Then, I used the general velocity relation to find the initial and final velocities based on the respective a value, the value for [tex]\mu[/tex] (mass of Mars times the Gravitational Constant, given in my textbook to be 4.293 E4 km^3/s^2), and the radius of Mars (3397 km).

I got an initial velocity of 4.43 km/s and a final velocity of 4.03 km/s, resulting in a [tex]\Delta V[/tex] of -0.4 km/s. I assume the negative value points to the fact that the delta V needs to be applied in the reverse direction to slow the vehicle down, resulting in a smaller apoapsis.

I just did this on a wing so I don't know if it is correct. Am I missing something here or does this seem correct?
 
Physics news on Phys.org
  • #2
Anyone?
 
  • #3
ColdFusion85 said:

Homework Equations


General Velocity Equation: V=[tex]\sqrt{{2\mu/R}-{\mu/a}}[/tex], where [tex]\mu[/tex]=GM, and a is the semi-major axis

Relation of radii to semi-major axis: [tex]r_p + r_a[/tex] = 2a, where [tex]r_p[/tex] is the radius at periapsis, and [tex]r_a[/tex] is the radius at apoapsis

This R is not the planetary radius. Think about it. Suppose you wanted to transfer from a circular orbit of radius a to a circular orbit of the same radius (i.e., no change in orbit). The delta-V for this maneuver is zero. You do not get zero if you use the planetary radius.
 
  • #4
yeah, R is just the distance from Mars
 
  • #5
The distance from the center of Mars, to be precise. The distance from Mars does involve the radius. It is the vehicle's altitude.
 
  • #6
D H said:
This R is not the planetary radius. Think about it. Suppose you wanted to transfer from a circular orbit of radius a to a circular orbit of the same radius (i.e., no change in orbit). The delta-V for this maneuver is zero. You do not get zero if you use the planetary radius.

I understand why the delta V would be zero if you did not want to change altitude, but why would you not get zero if you use the planetary radius? [tex]\mu[/tex] and the apoapsis would not change if you were at the same distance.

I am still confused as to what R I am supposed to use then. Is it the radius of Mars plus the distance to apoapsis? Or is it R the distance of Mars from the Sun?
 
  • #7
ColdFusion85 said:
I understand why the delta V would be zero if you did not want to change altitude, but why would you not get zero if you use the planetary radius? [tex]\mu[/tex] and the apoapsis would not change if you were at the same distance.

Yeah, I agree... don't see why delta V wouldn't be zero but...

I am still confused as to what R I am supposed to use then. Is it the radius of Mars plus the distance to apoapsis? Or is it R the distance of Mars from the Sun?

I think R is the distance of the orbiting body from the center of mass of the body being orbitted, as DH mentioned. ie... distance of the orbiting body from the center of mass of Mars.

In this case wouldn't it just be the periapsis radius?
 
Last edited:
  • #8
I used R= Radius of Mars + Distance to Periapsis. Is this correct? I figure if the delta V burn is going to be performed at the periapsis where the velocity vector is tangental to the radial component, the R at that point is the Radius to the surface of Mars (3397km) + the distance to the periapsis (3800km). Does this seem correct?
 
  • #9
ColdFusion85 said:
I used R= Radius of Mars + Distance to Periapsis. Is this correct? I figure if the delta V burn is going to be performed at the periapsis where the velocity vector is tangental to the radial component, the R at that point is the Radius to the surface of Mars (3397km) + the distance to the periapsis (3800km). Does this seem correct?

I thought the periapsis and apoapsis radius was measured from the center of mass of Mars... I'm not sure about this.
 
  • #10
You are correct. I just read that the periapsis is "the least distance of the elliptical orbit of an astronomical object from its center of attraction, which is generally the center of mass of the system."

So, I just use R=3800 km, and everything else looks alright?
 
  • #11
learningphysics said:
I thought the periapsis and apoapsis radius was measured from the center of mass of Mars... I'm not sure about this.

That's correct.

ColdFusion, if you are working with altitudes rather than distance from the center of the planet, everything you did in computing the eccentricity and semi-major axis is wrong. While apoapsis and periapsis sometimes are referenced with respect to planet surface, the semi-major axis never is. The equations you used for eccentricity implicitly assume planet-centered apoapsis/perapsis.

If these are indeed distances from the center of the planet, the only thing you did wrong was to add the planet radius to them.
 
  • #12
ColdFusion85 said:
You are correct. I just read that the periapsis is "the least distance of the elliptical orbit of an astronomical object from its center of attraction, which is generally the center of mass of the system."

So, I just use R=3800 km, and everything else looks alright?

I think so.. initial a = 7600km... final a = 4750km... R = 3800km... looks right.
 

1. What is an orbit transfer to lower apoapsis of a Mars orbit?

An orbit transfer to lower apoapsis of a Mars orbit is the process of changing the shape and altitude of a spacecraft's orbit around Mars in order to bring the apoapsis (the point farthest from the planet) closer to the surface of Mars.

2. Why would you want to transfer to a lower apoapsis in a Mars orbit?

Transferring to a lower apoapsis in a Mars orbit allows for a more efficient use of fuel and can help extend the lifespan of a spacecraft. It also allows for closer observation and data collection of the planet's surface and atmosphere.

3. What are the key considerations when planning an orbit transfer to lower apoapsis of a Mars orbit?

The key considerations when planning an orbit transfer to lower apoapsis of a Mars orbit include the amount of fuel needed, the timing and trajectory of the transfer, and the effects of Mars' gravity on the spacecraft during the transfer.

4. How is an orbit transfer to lower apoapsis of a Mars orbit executed?

An orbit transfer to lower apoapsis of a Mars orbit is typically executed by firing the spacecraft's engines in a specific direction and for a specific duration, calculated using orbital mechanics and the desired orbit parameters. This maneuver is often done in multiple stages to conserve fuel and adjust the spacecraft's trajectory as needed.

5. What challenges may arise during an orbit transfer to lower apoapsis of a Mars orbit?

Some challenges that may arise during an orbit transfer to lower apoapsis of a Mars orbit include unexpected changes in Mars' atmosphere, variations in the spacecraft's fuel consumption, and potential malfunctions or errors in the spacecraft's propulsion system. These challenges must be carefully monitored and addressed in order to successfully complete the orbit transfer.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
27
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
7K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
769
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
20K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top