Time of a falling object when the force of gravity isn't constant

[Shahar]

Homework Statement

Distance from planet = 10^14 meters, Radious of the planet = 10^7 meters, mass of the object = 100kg
g on the planet's surface = 10 m/s^2, g 10^14 meters away from the planets center = 2.5 m/s^2

Homework Equations

Fg=G*m*M/r^2

The Attempt at a Solution

I calculated the speed of the object at the surface of the planet ( 8931.5 m/s ) using Energy Conservation. I trןed to calculate the time using r(t), r(a) graphs.

 

[mfb]

g 10^14 meters away from the planets center = 2.5 m/s^2

Are you sure this is at 10¹⁴ meters instead of the surface? That would make the "planet" a supermassive black hole.

I calculated the speed of the object at the surface of the planet ( 8931.5 m/s ) using Energy Conservation

Can you calculate the speed at other distances, too?
How much time does the object need to get 1 meter (or in general, dx) closer, as function of distance?

 

[Shahar]

Are you sure this is at 10¹⁴ meters instead of the surface? That would make the "planet" a supermassive black hole.

It's 10 m/s^2 at the surface and 2.5 m/s^2 10^7 meters from the surface or 10^14 meters from the center.
I've calculated the planet's mass is 1.5*10^25 kg so its density is 35,000 km/m^3, Whitch is close to the density of the sun's core.

Can you calculate the speed at other distances, too?
How much time does the object need to get 1 meter (or in general, dx) closer, as function of distance?

And I used Energy Conservation to find the velocity equation.
V = √2 * √G*M/r2-G*M/r1

I can't find any function of time.

 

[haruspex]

It's 10 m/s^2 at the surface and 2.5 m/s^2 10^7 meters from the surface or 10^14 meters from the center.
I've calculated the planet's mass is 1.5*10^25 kg so its density is 35,000 km/m^3, Whitch is close to the density of the sun's core.
And I used Energy Conservation to find the velocity equation.
V = √2 * √G*M/r2-G*M/r1

I can't find any function of time.

10^14 is a lot more than twice 10^7.

 

[Shahar]

10^14 is a lot more than twice 10^7.

Oh right its not 10^14 its 2*10^14.

But I checked my calculation and I used the
right numbers there. So it 2.5 m/s^2 10^7 meters away from the surface and 10 m/s^2 at the surface . (This is a quote from the book).

But the data isn't the main thing. I want to find the v(t) equation.

 

[mfb]

Oh right its not 10^14 its 2*10^14.

Wait, what?
The density has another error of this type.

And I used Energy Conservation to find the velocity equation.
V = √2 * √G*M/r2-G*M/r1

Okay. v is the derivative of position with respect to time: v=dx/dt. What about 1/v? Can you see how to use that, especially if you combine it with the hint of post #2?

 

[Shahar]

Wait, what?
The density has another error of this type.Okay. v is the derivative of position with respect to time: v=dx/dt. What about 1/v? Can you see how to use that, especially if you combine it with the hint of post #2?

Sorry its 10^7 and 2*10^7. I'm really tired.

I think I get what you say.

 

[SteamKing]

It's 10 m/s^2 at the surface and 2.5 m/s^2 10^7 meters from the surface or 10^14 meters from the center.
I've calculated the planet's mass is 1.5*10^25 kg so its density is 35,000 km/m^3, Whitch is close to the density of the sun's core.

And I used Energy Conservation to find the velocity equation.
V = √2 * √G*M/r2-G*M/r1

I can't find any function of time.

Uhh, I'm not sure what kind of density 35,000 km/m³ is, but I'm pretty sure it's not the density at the sun's core.

That number is about 150 g / cm³, or say 150,000 kg/m³.

http://en.wikipedia.org/wiki/Sun

Pluto orbits at an average distance of 5.85×10¹² m from the sun, which is still far short of 10¹⁴ m.

 

[Shahar]

Uhh, I'm not sure what kind of density 35,000 km/m³ is, but I'm pretty sure it's not the density at the sun's core.

That number is about 150 g / cm³, or say 150,000 kg/m³.

http://en.wikipedia.org/wiki/Sun

Pluto orbits at an average distance of 5.85×10¹² m from the sun, which is still far short of 10¹⁴ m.

Yes, I know calculate all the data again with the correct numbers.

 

[Shahar]

OK, I calculated the planet's mass and the
object's velocity again. I got 1.5*10^25 kg for the planet's mass. And the velocity of the object is 10,000 m/s at the surface. I know 1414<t<2828 seconds.
I can't think of any equation that has time in it.
I did a(r) function and I found that the average acceleration is 5 m/s^2.
I used v(t) function for constant acceleration and I got 2,000 seconds. I'm not sure about that but I don't know anything about the time compared to something else.

 

[haruspex]

OK, I calculated the planet's mass and the
object's velocity again. I got 1.5*10^25 kg for the planet's mass. And the velocity of the object is 10,000 m/s at the surface. I know 1414<t<2828 seconds.
I can't think of any equation that has time in it.
I did a(r) function and I found that the average acceleration is 5 m/s^2.
I used v(t) function for constant acceleration and I got 2,000 seconds. I'm not sure about that but I don't know anything about the time compared to something else.

To get the time, you will need to go back to the differential equation of motion and solve it.

 

[Shahar]

What about 1/v?

I'm not sure but:
t=1/X'(v)
So I have to find the derivative of x(v).

EDIT : I tried that and it failed.

 

[HallsofIvy]

I presume you know that for gravity, $F= -\frac{GmM}{r^2}$. Since F= ma, $a= -\frac{GM}{r^2}$. Integrate to get the speed at time t and integrate again to get the position.

 

[Shahar]

I presume you know that for gravity, $F= -\frac{GmM}{r^2}$. Since F= ma, $a= -\frac{GM}{r^2}$. Integrate to get the speed at time t and integrate again to get the position.

Well, I know Newton law of gravtion but I don't know to integrate.
I tried every graph possible to get the v(t) but I can't find it.
I know the concept and basics of intgral calculus, but how I can get from the a(x) to the v(x)? In the last few hours I tried to do this with no secsuss.
I found v(x) and x(v).

 

[Chestermiller]

Have you learned about conservation of energy yet in your course?

Chet

 

[Shahar]

Have you learned about conservation of energy yet in your course?

Chet

Yes, and its just high school physics class.

 

[Chestermiller]

Yes, and its just high school physics class.

Do you know the equation for the potential energy of an object in a variable gravitational field (i.e., a gravitational field that varies with distance from a specified mass)?

Chet

 

[Shahar]

Do you know the equation for the potential energy of an object in a variable gravitational field (i.e., a gravitational field that varies with distance from a specified mass)?

Chet

Yes .
Pe=-G*M*m/r

 

[Chestermiller]

Yes .
Pe=-G*M*m/r

Excellent, except for a missing minus sign. So the sum of the potential energy and the kinetic energy of the falling body doesn't change as it falls. Let v = the velocity of the body at time t, and let r be the distance from the center of the planet at time t. What is the sum of the potential energy and the kinetic energy at time t? What is the sum at time zero? What is the velocity as a function of distance from the center?

Chet

 

[mfb]

Didn't we have that formula in post #3 already?

 

[Shahar]

Didn't we have that formula in post #3 already?

Yes , V(x).

What is the sum of the potential energy and the kinetic energy at time t?

Chet

The sum doesn't change due to Energy conversation. And in t=0 we only have potential energy.
I can't find a function that contains time.
Its constantly changing so I can't simply multiply it with other variables, can't I?
I tried to use averages and triangle area but it didn't work.

 

[mfb]

You'll need a differential equation. See my previous posts. Solving this differential equation is just an integral.

 

[Shahar]

You'll need a differential equation. See my previous posts. Solving this differential equation is just an integral.

I haven't learned yet how to solve a differential equation. I'm quite new to calculus.
But generally I need to take the derivative with respect to x . And this is the change in time,right ?

 

[mfb]

Change in time as function of distance is a good approach.

 

[Chestermiller]

Are trying to find the velocity when the body hits or the amount of time it takes to fall?

Chet

 

[Shahar]

Are trying to find the velocity when the body hits or the amount of time it takes to fall?

Chet

The time ,I already found the velocity.

 

[Chestermiller]

Didn't we have that formula in post #3 already?

Oops. Missed that. Got to be more careful.

The time ,I already found the velocity.

In post number 6, mfb gave you the trick. You know that v(r) = -dr/dt, so dt/dr = -1/v(r). So you have an equation for the derivative of t with respect to r in terms of r. (That's analogous to having dy/dx = f(x)). If you've had integration in your calculus course, then you should work on figuring out how to do this integration to get t. You also know that an integral can be represented graphically as the area under a curve, so if you plot dt/dr vs r, you can also get the integral by "counting boxes" on the graph.

Chet

 

[rcgldr]

V = √2 * √G*M/r2-G*M/r1
I can't find any function of time.

As mentioned, you'll need higher level math to determine when the objects collide. Alternate methods for determining velocity versus position:

$$a = \frac{d^{2}r}{dt^{2}}=-\frac{2GM}{r^{2}}$$
Method 1, multiply both sides by v = dr/dt:
$$\frac{d^{2}r}{dt^{2}}\frac{dr}{dt}=-\frac{2GM}{r^{2}}\frac{dr}{dt}$$
$$\frac{dv}{dt} \ v =-\frac{2GM}{r^{2}}\frac{dr}{dt}$$
integrate both sides:
$$\frac{1}{2}v^2 = \frac{GM}{r}-\frac{GM}{r_0}$$
Method 2, use chain rule (v as a function of r, r as a function of t, a = v'(r) r'(t))
$$\frac{dv}{dr}\frac{dr}{dt} = -\frac{2GM}{r^{2}}$$
$$dv \ v = -\frac{2GM}{r^{2}}{dr}$$
integrate both sides:
$$\frac{1}{2}v^2 = \frac{GM}{r}-\frac{GM}{r_0}$$
using - G M / r₀ as the constant of integration since v = 0 when r = r₀
$$v = dr/dt = -\sqrt{\frac{2 \ G \ M}{r} - \frac{2 \ G \ M}{r_0}} = -\sqrt{\frac{2 \ G \ M \ (r_0 \ - \ r)}{r_0 \ r}} = -\sqrt{\frac{2 \ G \ M}{r_0}}\sqrt{\frac{r_0 \ - \ r}{r}}$$
This basically matches the answer shown in post #3. To get time you'd have to integrate:
$$-\sqrt{\frac{r_0}{2 \ G \ M}}\sqrt{\frac{r}{r_0 \ - \ r}} \ dr = dt$$
The left side is complicated:

http://www.wolframalpha.com/input/?i=integral+-+sqrt(r0+/+(2+G+M))+sqrt(r+/+(r0-r))+dr

 

[Shahar]

As mentioned, you'll need higher level math.

Alternate approach for v(r):

v = dr/dt
a = dv/dt = (dv/dt) (dr/dr) = (dr/dt) (dv/dr) = - G M / r^2

v dv = - G M dr / r^2
1/2 v^2 = G M / r - G M / r0 (using GM / r0 as the constant of integration)
v = dr/dt = -sqrt ((2 G M / r) - (2 G M / r0))

To get time you'd have to integrate:

- dr / sqrt ((2 G M / r) - (2 G M / r0)) = dt

- sqrt(r r0) dr / sqrt(2 G M (r0 - r)) = dt

- sqrt(r0 / (2 G M)) sqrt(r / (r0-r)) dr = dt

As mentioned, you'll need higher level math.http://www.wolframalpha.com/input/?i=integral+-+sqrt(r0+/+(2+G+M))+sqrt(r+/+(r0-r))+dr

The answer in the site isn't very helpful.

 

[SammyS]

As mentioned, you'll need higher level math.

Alternate approach for v(r):

v = dr/dt
a = dv/dt = (dv/dt) (dr/dr) = (dr/dt) (dv/dr)

Really ?

(dv/dt) (dr/dr) = (dr/dt) (dv/dr) ?

The Calculus gods have just gone nuts !

Added in Edit: The RED emphasis above.

 

[Shahar]

Really ?

(dv/dt) (dr/dr) = (dr/dt) (dv/dr) ?

The Calculus gods have just gone nuts !

The link. Not this site.Sorry my english isn't that good.

 

[SammyS]

The link. Not this site.Sorry my english isn't that good.

@Shahar,
I was replying to Post # 28 by rcgldr .

 

[rcgldr]

The answer in the site isn't very helpful.

You can substitute u = sqrt(r / (r₀-r))

 

[Chestermiller]

I prefer the substitution r = r₀sin²θ

Chet

 

[rcgldr]

a = dv/dt = (dv/dt) (dr/dr) = (dr/dt) (dv/dr) = v dv/dr

Really ?

Chain rule, wiki link:
http://en.wikipedia.org/wiki/Chain_rule
I fixed what I now call method 1 and method 2 in my previous post.