(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

General Velocity Equation: V=[tex]\sqrt{{2\mu/R}-{\mu/a}}[/tex], where [tex]\mu[/tex]=GM, andais 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

3. 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 respectiveavalue, 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?

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# Homework Help: Orbit Transfer to Lower Apoapsis of a Mars Orbit

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