Calculate Earth's escape velocity using different methods

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Homework Help Overview

The discussion revolves around calculating Earth's escape velocity using various methods, including Newton's laws and the Lagrangian approach. Participants are exploring the relationships between variables involved in the equations of motion and the implications of initial conditions on the escape velocity calculation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using differential equations derived from Newton's laws and the Lagrangian to approach the problem. Questions arise regarding the relationship between radius and velocity, and how to derive a constant escape velocity from time-dependent functions.

Discussion Status

The discussion is active, with participants providing hints and exploring different methods. Some guidance has been offered regarding integrating equations and the significance of initial conditions, but no consensus or final solution has been reached.

Contextual Notes

Participants express uncertainty about the original problem statement, particularly regarding the requirement to use the Lagrangian equation for the calculation. There is also mention of the need for initial conditions to compute escape velocity effectively.

Physicsdudee
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Homework Statement
We were asked to calculate earth's escape velocity using methods 1) Newton law of motion, 2) Conservation of energy, 3) Lagrange method.
As to conservation of energy, that is clear, and I got the result (ca. 11.19km/s).
However for 1) and 3) I have no Idea, though I might have a start for 3).
Help would be appreciated.
Thanks
Relevant Equations
F=m*a, Ekin+Epot=const
try.PNG
 
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For 1) you could use Newton's laws to generate a differential equation and apply the relevant boundary conditions.

I must confess I'm not sure what is meant by 3). You can use the Lagrangian to generate either Newton's laws or the conservation of energy (which follows from time invariance of the Lagrangian).
 
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Note that the equation
$$m\dot v + \frac{GMm}{r^2}=0$$ is just ##F=ma##. You can't integrate it the way you did since ##r## is a function of ##t##.

How are ##r## and ##v## related?
 
Hmm I mean I know that r and v are related in a way that dr/dt=v and vice versa with the integral.
I could replace the v' in this differential equation you posted with r''.
That would give me a second order differential equation in terms of r, however, if I solve for r, I get a function r(t). How do I get a constant velocity out of this? By differentiating I get a function v(t), but I have no initial condotins to be able to compute v_escape.
 
Physicsdudee said:
Hmm I mean I know that r and v are related in a way that dr/dt=v and vice versa with the integral.
I could replace the v' in this differential equation you posted with r''.
That would give me a second order differential equation in terms of r, however, if I solve for r, I get a function r(t). How do I get a constant velocity out of this? By differentiating I get a function v(t), but I have no initial condotins to be able to compute v_escape.
I'm not sure what you are trying to do. Escape velocity is an initial condition at some initial ##r_0## in order to ensure that ##v \rightarrow 0## as ##r \rightarrow \infty##.
 
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Hmm yeah that makes sense. Tbh I don't really get the question we were given either. It said that we must calculate escape velocity using the Lagrange equation, that's it. But yeah, I will try to find a way, maybe some idea pops up.
 
Physicsdudee said:
Hmm I mean I know that r and v are related in a way that dr/dt=v and vice versa with the integral.
I could replace the v' in this differential equation you posted with r''.
That would give me a second order differential equation in terms of r, however, if I solve for r, I get a function r(t). How do I get a constant velocity out of this? By differentiating I get a function v(t), but I have no initial conditions to be able to compute v_escape.
The usual trick is to multiply both sides of the equation by ##\dot r##, which makes it easy to integrate. You should recognize the resulting equation.
 
vela said:
The usual trick is to multiply both sides of the equation by ##\dot r##, which makes it easy to integrate. You should recognize the resulting equation.
Thank you for the hint, I had totally forgotten about that trick. I think that made it work:)

try.PNG
 
As an aside comment @Physicsdudee -- please check out the "LaTeX Guide" link below the Edit window. Posting math equations in LaTeX makes them *much* more legible. Thanks.
 
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