Gravitational Attraction Changing with Distance

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Homework Statement


How would one go about calculating the increasing gravitational attraction between objects, one assumed for the purpose of the calculation to be motionless, and the other moving towards it according to the gravitational attraction.?



Homework Equations


F = GMm/r^2



The Attempt at a Solution



The above formula gives the gravitational attraction between objects at a set distance, but in order to know the attraction at multiple points one would have to go throught the entire equation for each new distance.

I would assume such a formula would involve the inverse square law, although any guess on my part. as to what that formula would be, would be just that, a guess.

Thank's in advance for your help.
 

Answers and Replies

  • #2
Doc Al
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Not sure what you are asking. That formula--Newton's Law of Universal Gravity--gives you the gravitational attraction between two (point) objects in terms of distance. So that is your answer. (And, yes, it's an inverse square law.)

Perhaps you can provide the exact problem you are trying to solve?
 
  • #3
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Not sure what you are asking. That formula--Newton's Law of Universal Gravity--gives you the gravitational attraction between two (point) objects in terms of distance. So that is your answer. (And, yes, it's an inverse square law.)

Perhaps you can provide the exact problem you are trying to solve?
Sure.

An object is placed so that it is allowed to move free of friction (basically getting as close to a vacuum as possible in your average home by having it suspended slightly above the ground (similar to some of the experiments originally used to measure G). Near the object is placed another, more massive (12.1kg) object (the distance is 10 cm). The gravitational attraction will cause the smaller object being suspended (100g or .1kg) to move towards the larger object.

What I need to do is calculate the time it should take the small object to reach the larger object, taking into account the increased force acting upon it the closer they get.

So far, I have calculated the initial force to be 0.0000000132944 N, which would give a time of 1 226 seconds. This figure, however, would be inaccurate because the object would accellerate as more force acted on it, thereby decreasing the time it would take to reach the other mass.
 
  • #4
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Let me guess, your doing http://www.fourmilab.ch/gravitation/foobar/" [Broken] aren't you?

Anyway, this is something I've also wanted to know for some time. One way i think you could do it is to use calculus but i really don't know the steps.
 
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  • #5
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Let me guess, your doing http://www.fourmilab.ch/gravitation/foobar/" [Broken] aren't you?

Anyway, this is something I've also wanted to know for some time. One way i think you could do it is to use calculus but i really don't know the steps.
Well yes, basically, but my explanation is shorter, albeit far more confusing...

Same thing here relating to Calculus, except I don't know ANY steps for it...
 
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  • #6
it sounds like you want the rate of change of the force with respect to the distance r

so...

you want to find

[tex]\frac{dF}{dr}[/tex]
 
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  • #7
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You'd need to:
-add a very small dt to t0
-calculate f from r
-calculate a from f (or go straight from r to a)
-calculate dv over dt and add it to v0
-calculate dr over dt and add it to r0
And then repeat this.
 
  • #8
30
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You'd need to:
-add a very small dt to t0
-calculate f from r
-calculate a from f (or go straight from r to a)
-calculate dv over dt and add it to v0
-calculate dr over dt and add it to r0
And then repeat this.
So what would e the formula for this, and what eaxactly do those variables (assuming they are, in fact, variables) mean?

I tried calculating this a different way, using the force at given points (I used every 2 centimeters). Would it be correct that the gravitational force is undefined, or would it be infinite. If it is undefined, how would I go about calculating this without any knowledge of calculus?

I could use calculus if completely neccessary, but I would need a step by step explanation...
 
  • #9
25
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t1 is the time
dt is delta time
r1 is range (or distance)
f is force
a is accelleration
v1 is velocity
dv is delta velocity
dr is delta range
r0 is the initial range
v0 is the initial velocity
t0 is the initial time
m1 is the mass of the movable block
m2 is th mass of the big block

For simple linear motion where the big block is immobile:
t1 = t0 + dt
f = g*m1*m2/r0^2
a = f/m1
dv = dt*a
v1 = v0 + dv
dr = dt*v1
r1 = r0 + dr
When you are done you put r1 in r0 and v1 in v0 and begin again.
dt should be as small as possible.

The calculating is best done by something like a spreadsheet.

Quick question: Do you think the mass of the mobile block matters?
 
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  • #10
30
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t1 is the time
dt is delta time
r1 is range (or distance)
f is force
a is accelleration
v1 is velocity
dv is delta velocity
dr is delta range
r0 is the initial range
v0 is the initial velocity
t0 is the initial time
m1 is the mass of the movable block
m2 is th mass of the big block

For simple linear motion where the big block is immobile:
t1 = t0 + dt
f = g*m1*m2/r0^2
a = f/m1
dv = dt*a
v1 = v0 + dv
dr = dt*v1
r1 = r0 + dr
When you are done you put r1 in r0 and v1 in v0 and begin again.
dt should be as small as possible.

The calculating is best done by something like a spreadsheet.

Quick question: Do you think the mass of the mobile block matters?
What do the "delta"'s mean?

What would the equation look like expressed in the format used in the example: F= Gmm/r^2 ?

I'm a bit confused... are your equations individual or to be done separately, and how do they relate?

Quick answer: Yes, at least I think...
 
  • #11
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Delta means change.
So delta velocity is the change of velocity.

There is one equation that calculates force.
The next calculation uses force to calculate accelleration.
You can make one equation out of the two.

The aim is to come up with two equations:
v1 = v0 something dt
r1 = r0 something dt

Quick question two: Why? What do you think would happen to m1 if you start combining the equations?
 
  • #12
30
0
t1 is the time
dt is delta time
r1 is range (or distance)
f is force
a is accelleration
v1 is velocity
dv is delta velocity
dr is delta range
r0 is the initial range
v0 is the initial velocity
t0 is the initial time
m1 is the mass of the movable block
m2 is th mass of the big block

For simple linear motion where the big block is immobile:
t1 = t0 + dt
f = g*m1*m2/r0^2
a = f/m1
dv = dt*a
v1 = v0 + dv
dr = dt*v1
r1 = r0 + dr
When you are done you put r1 in r0 and v1 in v0 and begin again.
dt should be as small as possible.

The calculating is best done by something like a spreadsheet.
Using this, I get...

t1 = t0 + dt ...?
f = g*m1*m2/r0^2 0.0000000132944 N
a = f/m1 1.32944 × 10^-7 m∕s^2
dv = dt*a ...?
v1 = v0 + dv ...?
dr = dt*v1 ...?
r1 = r0 + dr ...?

What would be the next step?
 
  • #13
111
0
I tried to use latex to explain some integration reasoning, but I don't know how to use latex and i need it.
 
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  • #14
30
0
I tried to use latex to explain some integration reasoning, but I don't know how to use latex and i need it.
You can always use the computer notation, Eg 2 squared = 2^2.

For that weird integral s thingy... well you can use italics, eg. f...close enough.

EDIT: Or you could look for the correct equations, then copy them... The tags are easy enough: [ tex] [ \tex] (no spaces in actual tags).

Putting the letter "G" in the tags, you get [tex]\G[\tex]

EDIT: Maybe I don't know what I'm doing either...
 
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  • #15
30
0
t1 is the time
dt is delta time
r1 is range (or distance)
f is force
a is accelleration
v1 is velocity
dv is delta velocity
dr is delta range
r0 is the initial range
v0 is the initial velocity
t0 is the initial time
m1 is the mass of the movable block
m2 is th mass of the big block

For simple linear motion where the big block is immobile:
t1 = t0 + dt
f = g*m1*m2/r0^2
a = f/m1
dv = dt*a
v1 = v0 + dv
dr = dt*v1
r1 = r0 + dr
When you are done you put r1 in r0 and v1 in v0 and begin again.
dt should be as small as possible.
But the end goal is to FIND the amout of time it takes... It doesn't make sense to give a value to time when time is what you want to find...
 
  • #16
30
0
Perhaps to restate my original question would be best: How do you calculate the amount of time it takes an object to travel towards another object, accounting for the inverse square law relation between distance and gravitational attraction?
 
  • #17
111
0
The answer using LRAM with 1000 divisions is 1236186.553 seconds. I'd love to know how to do this as an integral, but I can't figure it out
 
  • #18
30
0
The answer using LRAM with 1000 divisions is 1236186.553 seconds. I'd love to know how to do this as an integral, but I can't figure it out

What is LRAM, and is it freeware?
 
  • #19
111
0
No, LRAM is a mathematical method of calculating area under a curve; a estimate of an integral. Allow me to attempt to explain.

I divided the distance into many smaller units. So I had a distance of .1 m - .1999 m then .1999 m - .1998 m and so on until i reached .0001 m

I then calculated the acceleration using the first distance in the fomula

A = G(m+M)/r

I set Vo for the first interval to zero and using v^2 = Vo^2 + 2ar I found the velocity at the end of the interval. (This would be done for every interval, using the final velocity of the previous interval as the initial velocity)

I could use the final velociy, initial velocity, and the acceleration to find the time spent in that interval. Adding all 1000 of these times gave me 1231.1 seconds.

This time is most definatly wrong, but it's in my calculations, the method seems correct to me.

I wish that someone would explain how to do this using integrals because i't bothering me.
 
  • #20
30
0
No, LRAM is a mathematical method of calculating area under a curve; a estimate of an integral. Allow me to attempt to explain.

I divided the distance into many smaller units. So I had a distance of .1 m - .1999 m then .1999 m - .1998 m and so on until i reached .0001 m

I then calculated the acceleration using the first distance in the fomula

A = G(m+M)/r

I set Vo for the first interval to zero and using v^2 = Vo^2 + 2ar I found the velocity at the end of the interval. (This would be done for every interval, using the final velocity of the previous interval as the initial velocity)

I could use the final velociy, initial velocity, and the acceleration to find the time spent in that interval. Adding all 1000 of these times gave me 1231.1 seconds.

This time is most definatly wrong, but it's in my calculations, the method seems correct to me.

I wish that someone would explain how to do this using integrals because i't bothering me.
My calculations gave 1 226 seconds, so whatever we're doing wrong appears to be similar...
 

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