- #1

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So say a body is released from rest a distance Ro from a point mass, and it's initial acceleration is that stated above- how would I go about finding the time taken to travel this distance (from Ro to the origin.

Thanks.

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- Thread starter jmz34
- Start date

- #1

- 29

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So say a body is released from rest a distance Ro from a point mass, and it's initial acceleration is that stated above- how would I go about finding the time taken to travel this distance (from Ro to the origin.

Thanks.

- #2

tiny-tim

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hi jmz34!

so r'' = -GM/r^{2} ?

multiply both sides by r' …

r'r'' = -GMr'/r^{2}

so r'' = -GM/r

multiply both sides by r' …

r'r'' = -GMr'/r

… and integrate

- #3

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hi jmz34!

so r'' = -GM/r^{2}?

multiply both sides by r' …

r'r'' = -GMr'/r^{2}

… and integrate

Using this method I did this:

d/dt(0.5*(r')^2)=(-GM/r^2)r'

then integrated once and simplified to get

(dr/dt)^2=2GM/r

solving this for t gives:

t=(1/3)*SQRT(2/GM)*Ro^(3/2)

If you could have a quick look at my method I'd be very grateful.

Thanks alot.

- #4

tiny-tim

Science Advisor

Homework Helper

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Using this method I did this:

d/dt(0.5*(r')^2)=(-GM/r^2)r'

then integrated once and simplified to get

(dr/dt)^2=2GM/r.

yes, but after that i get a bit lost …

you seem to have lost r completely, and you don't have a constant of integration

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