Solve Max Velocity Increase of 100kg Probe in Gravity Assist from Mars

[zebra1707]

Homework Statement

I have a question which asks for the maximum increase in velocity that can be achieved by a 100 kg space probe traveling at 12 000 ms -1, in a gravity assist manoeuvre around the planet Mars, which has an orbital velocity of 24200 ms -1.

Homework Equations

vf = vi + 2vi
= 12000 + 2(24200)
= 60400, therefore max increase in velocity is 60400 ms -1

The Attempt at a Solution

See above. The next part asks how would your answer be if the probe only completed 120 degree arc in its flyby? Support with vector diagram.

You are not told what the original arc is, so I assume it is greater than 120 degrees. However, that does not assist with the answer to this problem?

If I draw the Before and After vectors diagrams and resultant incoming and outgoing velocities relative to the planet by using pythag - but I am still no closer to a solution.

 

[Filip Larsen]

You have gotten the speeds mixed up. First note, that the speeds are (I assume) given relative to the sun and that the speeds in the first equation should be relative to the planet (mars), so you should transform the speed for the probe to be relative to the planet (by coincidence this also gives 12 km/s). Second, the initial and final speed in the equation refer to that of the probe, not the planet. Third, the text ask for the change in speed (vf minus vi).

If you are given an arc (angle between initial and final velocity vector) you also need another formula that contains this angle (or you need to derive it from geometry), but its difficult to provide help without knowing a bit more about what you have been given.

 

[zebra1707]

The question is as stated no additional information.

Formula comes from
1. Applying the conservation of momentum to this collision KVi - vi = KVf + vf
2. Applying the conservation of kinetic energy KVi^2 + vi^2 = KVf^2 + vf^2

Solving equations simultaneously for vf leads to the following equation:

vf = vi + 2Vi

where: vf = maximum exit velocity of spacecraft ms -1
vi = entry velocity of spacecraft - relative to the sun (12000 ms -1)
Vi = velocity of the planet (in this case Mars - 24 200 ms -1)

Therefore vf = 12,000 + 2(24 200)
= 60 400 ms -1

Therefore this is the max increase in velocity is 60 400 ms -1. The question then asks what would the answer be if the probe only completed 120 degree of arc in this fly by (support with vector diagram).

 

[Filip Larsen]

OK, ignoring for a minute how mind-numbing unrealistic this "collision" is, the change in speed in first case must be vf-(-vi), that is, a tad more than vf.

To calculate how much speed changes if the initial velocity is only turned 60 degrees (instead of 180), you could reformulate the momentum equation by looking only at the momentum projected onto one particular direction and then solve this equation together with the kinetic energy equation as before. For instance, if you draw the velocities before and after in a diagram with the planet velocities parallel to the x-axis, you could write an equation for conservation of momentum onto the x-axis. You can probably apply some assumptions along the way to simplify the calculations. Alternatively, you could look up how to calculate the final speeds in a two-dimensional elastic collision and apply this to your problem.

I must admit that I originally thought the problem text were concerned with real gravitational sling-shot maneuvers. If it were, it would be instructive to know that the maximum change in speed for a passive Mars gravity assist is around 3.5 km/s, and can be calculated as

$$\Delta v = 2 v_\infty \sin(\varphi / 2)$$

where $v_\infty$ is the hyperbolic excess speed and $\varphi$ is the rotation of the velocity vector which depends on the minimum distance the probe passes by the planet center (this is the equation I thought you referred to originally, just with the rotation angle set to 180 deg). Setting the minimum distance to the planet radius for Mars and finding the hyperbolic excess speed which maximizes change of speed gives the mentioned value of 3.5 km/s for mars.

 

[zebra1707]

Thank you Filip for your assistance. Now this makes sense.

CHeers P