# Help with Gravity Assist [closed]

In our scenario, a spaceship of mass 70,000 kg and velocity $$v_o$$ is planning on slingshotting around Earth and we're supposed to find what the exit velocity will be. The Earth is the origin in our coordinate plane and the system consists only of the Earth and the spaceship and no external force acts on the system. The right and up are positive while left and down are negative. The Earth's equator is assumed to be rotating at 0.00007 rad/sec or 460 m/s with respect to the planet's center and the planet's linear velocity is 0 m/s. Here is the conservation of linear momentum:

$$M_{earth}(460)\;+\;m_{spaceship}(v_o)\,=\,M_{earth}(v_{earth})\;+\;m_{spaceship}(-v_f)$$

The problem is, I don't know how to solve for the change in the planet's velocity or what to do now. Even if I do assume that the relative velocity of the Earth remains the same due to $$M_{earth}\gg m_{spaceship}$$, then the equation will turn out to be:

$$m_{spaceship}(v_o)\,=\,m_{spaceship}(-v_f)$$ $$v_f=-v_o$$

Then, the equation will imply that the spaceship is not given a gravity assist and the speed doesn't change. Am I doing something wrong or is there another way to do this? I need help understanding the math behind it on an intuitive level. Most websites or sources I go to have insufficient or too complicated of an explanation. Any help would be appreciated, thank you.

Ignore the letters on the image for now

## closed as off-topic by G. Smith, John Rennie, Jon Custer, GiorgioP, tpg2114♦May 18 at 1:00

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The insight that might be missing here is that this is essentially an elastic collision between two moving objects. That is because in your system, momentum is not only conserved, but so is Energy.

The other insight is that the final speed of the larger object is not exactly U it is simply so close to it that you are given that, but if you were to actually calculate the exact values using their masses then you would find that the difference makes up for the "missing" momentum and energy.

Finally in order to arrive at the above solution you must find the solution for the system:

$$$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$$$ $$\tfrac{1}{2}m_1v_{1i}^2 + \tfrac{1}{2}m_2v_{2i}^2 = \tfrac{1}{2}m_1v_{1f}^2 + \tfrac{1}{2}m_2v_{2f}^2$$

Solving these we get that the basic equation for a 1d elastic collision for object 1 with object 2 is:

$$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} + \frac{2m_2}{m_1+m_2}v_{2i}$$

The solution for object 2 is simply given by flipping the numbers.

Using what you described as $$M_{\text{earth}} \gg M_{\text{spaceship}}$$ we get that their limits are:

$$v_{1f} \approx 2v_{2i} - v_{1i} = -2U - v$$

$$v_{2f} \approx v_{2i} = -U$$

Notice again, $$\approx$$ not $$=$$, in reality the numbers are not those, they are just very close to the point where they have no relevance in this particular calculation (at least that is how the makers of that diagram think).

• Also an interesting shortcut that arises from this system is that sum of the initial and final velocity of object 1 must equal the sum of the initial and final velocity of object 2. This is usually described as "the speed of approach is also the speed of separation" and it is true for any 1d elastic collision. – ElectrumHafnium May 16 at 5:23
• Why is the planet's speed $U$ in the picture pointing to the left when the planet should be stationary and spinning? Then shouldn't you use $\frac{1}{2} I_{Earth} \omega ^2$? – nidarshans May 16 at 17:44
• well for one if U is 0 then that simply means that the U in the final velocities is also 0. Second of all the rate at which the planet spins is almost entirely negligible since the spaceship is presumably too small to feel tidal effects and the earth is way too large to slow it's rotation from the slingshot. If you wanted to calculate the change in rotational velocity you would need to find the terms for angular momentum of the earth-spaceship system when the spaceship is at point (0, $r_{orbit}$) and then when the angular momentum when the spaceship is at (0, -$r_{orbit}$). – ElectrumHafnium May 16 at 23:34
• Then you would use the fact that those two should be equal. – ElectrumHafnium May 16 at 23:35
• Here's a useful animation illustrating what I am describing. According to the planet providing the relative speed of the object doesn't change. But according to the Sun's frame the change is much more dramatic. – ElectrumHafnium May 17 at 2:02