Why Is Calculating Escape Velocity Different from Determining Orbital Speed?

In summary, the conversation discusses the equation for terminal velocity and its application to understanding planetary orbits. It also touches on the difference between escape velocity and the velocity needed to keep a test particle in circular orbit. One person is seeking help in finding their mistake in calculating the equation, while the other offers some corrections and clarifications. Overall, the conversation highlights the complexity and nuances involved in understanding planetary orbits and the importance of accurately applying mathematical concepts.
  • #1
realitybugll
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I tried finding the equation for terminal velocity in a broader attempt to understand planetary orbits better. According to wikipedia's two derivations (one of which i don't understand very well) I'm off by a factor of [tex] \sqrt 2 [/tex]. I would greatly appreciate help in finding my thinking/math mistake. Thanks in advance!

I was reading another thread on the same topic, and I think what I have calculated is the "velocity needed to keep the test particle in circular orbit of radius r" which is apparently different than "escape velocity".

I started off assuming that the planet is questions velocity vector is perpendicular to the force one, attracting it to the larger body whose movement in relation to the planet is negligible. So we should be talking about a circular orbit.

I thought that for every miniscule displacement towards the larger body there has to be a simultaneous one perpendicular such that the distance between the objects remains constant.

Mathematically, [tex] {r - d_1}^{2} + d_2^{2} = r^{2} [/tex]

where [tex] d_1 [/tex] is the displacement towards the larger body, [tex] d_2 [/tex] is the displacement perpendicular to that resulting from the planets inherent velocity, and [tex] r [/tex] is the radius or displacement between the objects.

[tex] d_1 [/tex] can be substituted with [tex] \frac {at^2} {2} [/tex] where a is the acceleration of the planet towards the larger body and t the time.

similarly [tex] d_1 [/tex] can be substituted with [tex] vt [/tex] where v is the planets escape velocity and t the time.

So rearranging and solving for v we get:

[tex] d_1 = \sqrt {2rd_2 - d_2^{2}} [/tex] [tex] v = \frac {\sqrt {rat^{2} - {\frac {a^{2}t^{4}} {4}}}} {t} [/tex]

and taking the limit of t, the time, as it tends to zero and thus minimizes the two theoretical displacements of the planet (that from the acceleration due the gravity of the larger planet and the perpendicular motion from its own velocity) making the curvature of the orbit more continuous and closer to circular, results in I think (I haven't really learned about limits)

[tex] v = \sqrt {ra} [/tex] whereas it really is [tex] v = \sqrt {2ra} [/tex]

note: I was reading another thread on the same topic, and I think what I have calculated is the "velocity needed to keep the test particle in circular orbit of radius r" which is apparently different than "escape velocity". I had been thinking they were the same, but I guess escape velocity refers to the velocity when elliptical orbits can't form (?).
 
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  • #2
Yes, you do not understand escape velocity. Look it up. Also, saying that ##d_1=\frac{1}{2}at^2## is nonsense. It holds only when the acceleration of the mass is constant which is not the case here. It looks like your understanding of planetary orbits was limited when you posted this 9 years ago. I hope it has improved since then.
 

1. What is escape velocity?

Escape velocity is the minimum speed that an object needs to achieve in order to overcome the gravitational pull of a larger object and escape its orbit.

2. How is escape velocity derived?

Escape velocity is derived using the equation: Ve = √(2GM/r), where G is the gravitational constant, M is the mass of the larger object, and r is the distance between the objects.

3. Why is escape velocity important?

Escape velocity is important in understanding space travel and the ability of objects to escape the gravitational pull of planets and other celestial bodies. It also helps us understand the behavior of objects in orbit.

4. How does escape velocity differ on different planets?

Escape velocity differs on different planets because it is dependent on the mass and size of the planet. For example, the escape velocity on Earth is 11.2 km/s, while on the Moon it is only 2.38 km/s.

5. Can escape velocity be exceeded?

Yes, it is possible for an object to exceed escape velocity. This is known as hyperbolic excess velocity and occurs when an object has enough speed to escape the gravitational pull of a larger object, but not enough to leave the entire system completely.

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