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Finding time required via force equation

  • Thread starter shanepitts
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The problem, relevant equations, and my attempt at a solution are all shown on both attached images.
Not sure where the π and and extra (1/4)1/2 is coming from.
Also, I noticed that my final result, in the attemp at a solution should be t=(mb3/4k)1/2, no negative sign.


image.jpg
image.jpg
 
Last edited:

haruspex

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Your velocity calculation is wrong. That definite integral will give the velocity at the origin (which will be infinite), but for the next step you need the velocity at an arbitrary distance x.
 
84
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Your velocity calculation is wrong. That definite integral will give the velocity at the origin (which will be infinite), but for the next step you need the velocity at an arbitrary distance x.
Thanks for the quick reply, but not to seem to ignorant in relation to using definite integrals, why would the velocity be infinite at origin?
 

haruspex

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Thanks for the quick reply, but not to seem to ignorant in relation to using definite integrals, why would the velocity be infinite at origin?
What did the indefinite integral look like? What happened when you applied the x=0 bound?
 
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What did the indefinite integral look like? What happened when you applied the x=0 bound?
so obvious, it gets divided by zero; it Is undefined.

Thanks again
 

ehild

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Such problem can be solved by using Kepler's Third Law. This is planetary motion, only the ellipse is very-very elongated. Still, the time of revolution is the same as along an equivalent circle, with radius equal to the semi-mayor axis of this orbit.
From the definition of ellipse, the sum of the distances between the planet and the foci is equal to the mayor axis, 2a. And it is the same as ##2\sqrt{f^2+b^2} where b is half of the minor axis.
If you make the ellipse narrower and narrower, at the limit of b=0, you get that a=f. The semi-mayor axis is equal to the distance of a focus from the centre. The planet starts from one focus and arrives to the Sun at the other focus in half of the time period.
According to Newton's Third Law, the time period is the same as that on a circle, with the same radius as the semi-mayor axis of the distorted ellipse.
The initial distance between planet and Sun is d = 2a. What is the time period along a circular orbit with radius d/2?


fallingtime.JPG
 

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