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Object falling towards moon

  • Thread starter dbogen
  • Start date
  • #1
2
0

Homework Statement


An object is dropped from a distanse of 2*R(moon radius) from moon senter.
How many seconds does it take until impact with moon, and in what speed will it hit?

The distanse of the fall will be 1.74*10^6 m. The problem is the not so constant accleration.. ;)

gravity constant: G := 6.67*10^(-11)
moon mass M := 0.0735^24 kg
moon radius R := 1.74*10^6 m



Homework Equations



gravity(accleration) is
g=G*M/R^2

So a(x)=(G*M)/(2*R - x)^2 , x = meter fallen x{0..R)


The Attempt at a Solution


I've calculated the speed at impact:

Average accleration:
> A:=(G*M)/(2*R - x)^2
> Aa := (int(A, x = 0 .. R))/R;
0.8096264368

Time using x= 0.5*a*t^2
> T := solve(R = 0.5*Aa*t^2, t);
-2073.229031, 2073.229031

This time is found using constant accleration, so it isnt the correct one...
but it will work for finding the speed of impact.

Speed
> V := T*Aa;
-1678.541033, 1678.541033
So it will hit at 1678,54 m/s

but for how long will it fall?
 

Answers and Replies

  • #2
430
2
You have acceleration as a function of distance. Put it in the expression a = v*dv/dx.That way you can obtain velocity as a function of distance by integrating. That will solve the first part of your problem. For time you can write v as dx/dt. Now if you put that and integrate you have distance as a function of time.
 
  • #3
2
0
Thanks aim!

I got the correct velocity using your method, but i dont figure out the distanse as a function of time...

Could you please try to explain it again, a bit more detailed? :)
 
  • #4
430
2
Ok you figured out velocity as a function of distance(x). Now what you do is write v as dx/dt. You will notice that the equation now contains x and t only. A little rearrangement and integration and you can express x as function of t.
 

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