Acceleration as a function of position, and time taken to travel a distance

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SUMMARY

The discussion focuses on calculating the time taken for a body to travel a distance under the influence of gravitational acceleration, defined as a=-GM/(Ro^2). The user starts from rest and employs the equation r'' = -GM/r^2, integrating to derive (dr/dt)^2 = 2GM/r. The final expression for time, t, is determined as t = (1/3) * SQRT(2/GM) * Ro^(3/2). The conversation also highlights the importance of including a constant of integration in the calculations.

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jmz34
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I've been doing a problem that requires me to find the time taken to travel a certain distance if I know the initial acceleration of a body at the starting position and its initial velocity (starts from rest). The acceleration is a function of position a=-GM/(Ro^2).

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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hi jmz34! :wink:

so r'' = -GM/r2 ?

multiply both sides by r' …

r'r'' = -GMr'/r2

… and integrate :smile:
 
tiny-tim said:
hi jmz34! :wink:

so r'' = -GM/r2 ?

multiply both sides by r' …

r'r'' = -GMr'/r2

… and integrate :smile:

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 a lot.
 
jmz34 said:
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 :confused:
 

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