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

[SammyS]

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

Chain rule, wiki link:

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

This is used when acceleration is a function of distance (instead of time), acceleration = a(r) = - G M / r^2.

I'm familiar with the chain rule.

It's the abuse of the notation in that set of steps that I was objecting to, in particular the (dv/dt)(dr/dr) in

dv/dt = (dv/dt) (dr/dr) = (dr/dt) (dv/dr) .​

dv/dt is not a fraction. It represents a derivative in Leibniz's notation . It is very handy for implementing the chain rule.

You don't simply take $\displaystyle\ \frac{dv}{dt}\,,\$ then multiply numerator & denominator by $\ dr\$ like so, $\displaystyle\ \frac{dv}{dt}\frac{dr}{dr}\$ and then swap locations of dr & dt or dv to get $\displaystyle\ \frac{dv}{dr}\cdot\frac{dr}{dt}\$.

If v is a function of r, and in turn, if r can be expressed as a function of t, then $\displaystyle\ \frac{dv}{dt}=v'(r)\cdot r'(t)=\frac{dv}{dr}\cdot\frac{dr}{dt}\$.

 

[Chestermiller]

Maybe if I had paid attention to the thread title ... (brain fade on my part). In that case, can you delete all of the posts with the solutions? Or if you give me an OK, I can go back and empty out all my previous posts, leaving a comment about waiting for feedback. Perhaps the OP hasn't visited this thread since those posts were added.

Good suggestion. I will go back and delete and edit where appropriate. In the Country where the OP lives, it is a little before sunrise now, so maybe he hasn't seen any of the posts yet.

Chet

 

[Shahar]

Update :I am tring to solve this using basoc calculus. So I'm tring to find the a(t) using a(x). Because v'(t)=a(t). I'm a bit stuck but I think I'm able to solve it.

 

[mfb]

Converting a(x) to a(t) is difficult, and requires finding the time (and everything else) first. The correct approach is in the thread now, you just have to integrate it. If you don't know how to calculate integrals then I don't see a way to solve the problem.

 

[Shahar]

Converting a(x) to a(t) is difficult, and requires finding the time (and everything else) first. The correct approach is in the thread now, you just have to integrate it. If you don't know how to calculate integrals then I don't see a way to solve the problem.

II integrated a(x) and it gave me , well...IIts not the average velocity.

Converting a(x) to a(t) is diffi

Converting a(x) to a(t) is difficult, and requires finding the time (and everything else) first. The correct approach is in the thread now, you just have to integrate it. If you don't know how to calculate integrals then I don't see a way to solve the problem.

I integrated a(x) and v(x). The big problem is that I don't really understand differentials.

 

[Chestermiller]

II integrated a(x) and it gave me , well...IIts not the average velocity.
I integrated a(x) and v(x). The big problem is that I don't really understand differentials.

If you don't understand differentials, how can you say that you have done an integration? Have you learned how to do integration in your courses yet? Have you learned about doing integrations by using algebraic or trigonometric substitution?

Chet

 

[Shahar]

If you don't understand differentials, how can you say that you have done an integration? Have you learned how to do integration in your courses yet? Have you learned about doing integrations by using algebraic or trigonometric substitution?

Chet

Because I can't solve this problem without knowing how to integrate I learned integrations using algebraic substitution.

 

[Chestermiller]

What about doing it graphically? Is that allowed?

Chet

 

[mfb]

Don't integrate v(x), integrate 1/v(x).

 

[Shahar]

What about doing it graphically? Is that allowed?

Chet

My question is not homework. It's too advance to be 10th grade homework. In the original question they asked us to find a time range for the falling object .(1414<t<2828 seconds) .
I just wanted to find the precise time it takes to an object moving with changing gravitational acceloration to get from A to B.

Don't integrate v(x), integrate 1/v(x).

I tried that,

 

[rcgldr]

This problem starts off with an equation for acceleration as a function of position. The sequence here is to generate an equation for velocity as a function of position, then time as a function of position, then invert this to an equation for position as a function of time, take the derivative to get velocity as a function of time, and another derivative to get acceleration as a function of time. However at some point in this process, you may not be able to continue, either due to an equation that can't be integrated, or an equation for time as a function of position that can't be inverted to an equation for position as a function of time.

If you wan't to get an idea of how this process works, try a simpler case like a(x) = -x (acceleration as a function of position), with an initial position of 0 and initial velocity of 1.

 

[Shahar]

This problem starts off with an equation for acceleration as a function of position. The sequence here is to generate an equation for velocity as a function of position, then time as a function of position, then invert this to an equation for position as a function of time, take the derivative to get velocity as a function of time, and another derivative to get acceleration as a function of time. However at some point in this process, you may not be able to continue, either due to an equation that can't be integrated, or an equation for time as a function of position that can't be inverted to an equation for position as a function of time.

If you wan't to get an idea of how this process works, try a simpler case like a(x) = -x (acceleration as a function of position), with an initial position of 0 and initial velocity of 1.

This is what I'm doing right now.
It will take some time.

 

[rcgldr]

My question is not homework. I just wanted to find the precise time it takes to an object moving with changing gravitational acceloration to get from A to B.

Well you already have the solution for time versus position in post #28 (the link to the math site), but in the form of a very complicated equation resulting from integration of velocity versus position. update - the rest of this post about substitutions was moved to post #50 to combine the important equations into a single post.

 

[Shahar]

This problem starts off with an equation for acceleration as a function of position. The sequence here is to generate an equation for velocity as a function of position, then time as a function of position, then invert this to an equation for position as a function of time, take the derivative to get velocity as a function of time, and another derivative to get acceleration as a function of time. However at some point in this process, you may not be able to continue, either due to an equation that can't be integrated, or an equation for time as a function of position that can't be inverted to an equation for position as a function of time.

If you wan't to get an idea of how this process works, try a simpler case like a(x) = -x (acceleration as a function of position), with an initial position of 0 and initial velocity of 1.

Yeah, its impossible to get v(t). I'm now just searching for x(t).
I know there are a lot of solutions here but I don't really understand them. I think I have to get to them myself to fully understand this.

 

[rcgldr]

I'm now just searching for x(t). I know there are a lot of solutions here but I don't really understand them. I think I have to get to them myself to fully understand this.

You need to get t(r) first, by integrating the equation you got in post #3, which I redid in post #28.
The left term, - sqrt(r₀ / (2 G M)) is a constant, so the issue here is integrating the right term, sqrt(r / (r₀ - r)) dr .
r₀ is the initial position, in this case 10^14 meters.

$$-\sqrt{\frac{r_0}{2 \ G \ M}}\sqrt{\frac{r}{r_0 \ - \ r}} \ dr = dt$$
The substitutions suggested in posts #33 and #34 will result in a much simpler equation, but you'll have to convert position into the substituted values to order to calculate the time.

post #33 substitution (r₀ is a constant):
$$ u = \sqrt{\frac{r}{r_0 \ - \ r}} $$
$$r = r_0 - \frac{r_0}{1 + u^2} $$
post #34 substitution (this one is simpler):
$$ r = r_0 \ sin^2(θ) $$
$$θ = {sin^{-1}\left ( \sqrt{\frac{r}{r_0}} \right )} $$
You need to determine what dr (derivative of r) is when using either substitution.

Use the post #34 substitution which is simpler than the post #33 substitution. The resulting integrals with either substitution are complicated enough that you'll probably want to look up the solution using a table of integrals (old timers may remember the CRC books with tables of integrals) or a website like WolframAlpha to solve, but those resulting integrals will be much simpler than the one from post #28.

 

[Chestermiller]

Hi Shahar,

Are you still out there, and are you still interested in pursuing the solution to this problem? Is anyone else out there interested in having rcgldr help them do this integration? rcgldr would like to complete the solution so that he does not have to save it in his computer files. If I don't hear back from someone on this by Monday, I'm going to release rcgldr to provide his analysis. Then I'm going to close this thread.

Chet

 

[Shahar]

Hi Shahar,

Are you still out there, and are you still interested in pursuing the solution to this problem? Is anyone else out there interested in having rcgldr help them do this integration? rcgldr would like to complete the solution so that he does not have to save it in his computer files. If I don't hear back from someone on this by Monday, I'm going to release rcgldr to provide his analysis. Then I'm going to close this thread.

Chet

I found a way to get the time.But because I have to learn more calculus I can't derive an X(t) function.
I'm learning calculus alone amd when I will understand how to solve a deferential equation Ill try to derive an X(t) function.
So I think rcgldr can release his analysis now.

 

[Chestermiller]

I found a way to get the time.But because I have to learn more calculus I can't derive an X(t) function.
So I think rcgldr can release his analysis now.

Thanks. Personally, I don't see how you could get the time without automatically also getting X(t) as a byproduct, but that's just me.

OK, regldr. You're free to proceed.

Chet

 

[Shahar]

Thanks. Personally, I don't see how you could get the time without automatically also getting X(t) as a byproduct, but that's just me.

OK, regldr. You're free to proceed.

Chet

I used averages(V , a) .
If I just combine all of the steps into one big expression, would it be an X(t) equation?

 

[Chestermiller]

I used averages(V , a) .
If I just combine all of the steps into one big expression, would it be an X(t) equation?

Why don't you show us what you did and how you arrived at your final answer? Then we can compare it with what rcgldr got by integration.

Chet

 

[Shahar]

Using conversation of energy I found V(r) => Vfinal.
And I know the function a(r) = G × M / r2.
So when V0 = 0 ,∫ a(r) dr = (Vfinal * Δt)*(Δt/Vavg) = Vfinal * Vavg.
Δt = Δx/Vavg.

 

[Chestermiller]

Using conversation of energy I found V(r) => Vfinal.
And I know the function a(r) = G × M / r2.
So when V0 = 0 ,∫ a(r) dr = (Vfinal * Δt)*(Δt/Vavg) = Vfinal * Vavg.
Δt = Δx/Vavg.

OK. What you basically did here was to use a very crude approximation to the average velocity, equal to half the final velocity. This would not give a very accurate result for Δt. Please tell us what your result for Δt was so that we can compare it with what rcgldr got by doing the integration exactly.

Chet

 

[Shahar]

OK. What you basically did here was to use a very crude approximation to the average velocity, equal to half the final velocity. This would not give a very accurate result for Δt. Please tell us what your result for Δt was so that we can compare it with what rcgldr got by doing the integration exactly.

Chet

I got 2000 seconds, which is the only answer that is in the limit 1414 < t < 2828 seconds.
I found that the average accelerationis 5 m/s. a0 = 2.5 m/s2 and af = 10 m/s2.
I rejected this idea twice already but becuse it's the on;y one that gives an answer in the limit I tried it again.
The answer is a bit too small, isn't it?

EDIT: I can't find any material about differentials in physics. Most of the stuff on the internet is a bit too complex for me.
Do you know a simple explanation for differentials in physics?

 

[Chestermiller]

I got 2000 seconds, which is the only answer that is in the limit 1414 < t < 2828 seconds.
I found that the average accelerationis 5 m/s. a0 = 2.5 m/s2 and af = 10 m/s2.
I rejected this idea twice already but becuse it's the on;y one that gives an answer in the limit I tried it again.
The answer is a bit too small, isn't it?

EDIT: I can't find any material about differentials in physics. Most of the stuff on the internet is a bit too complex for me.
Do you know a simple explanation for differentials in physics?

In post #45, you indicated that you had tried plotting 1/v as a function of r, and integrating graphically to get the area under the curve, but that didn't work. Can you elaborate on this? Can you furnish the plot? We would like to find out why it didn't work for you. (It should have).

Chet

 

[Shahar]

In post #45, you indicated that you had tried plotting 1/v as a function of r, and integrating graphically to get the area under the curve, but that didn't work. Can you elaborate on this? Can you furnish the plot? We would like to find out why it didn't work for you. (It should have).

Chet

I integrated 1/v(r) dr (and v(r) )and I had an expression that I cna't understand what it means.

 

[Chestermiller]

I integrated 1/v(r) dr (and v(r) )and I had an expression that I cna't understand what it means.

Let's see what you did. This is exactly what rdgldr did, so we can compare notes, and, if you went wrong, he can point out where.

Chet

 

[Shahar]

a = G * M
v(x) = √(2 *a)/x
1/v(x) = 1/√(2 *a)/x = √x / √2a

∫ 1/v(x) dx =
=∫ √x / √2a dx =
=1/√2a ∫ √x =
=1/√2a * x^1.5/1.5 =
= √2 * x^1.5/3a

I'm not sure it's correct.
And I reread post #27 about this equation but I don't undersatnd how to apply it.
From what I understand this is t, and It doesn't fit when I plug in x1 and x2.

 

[Chestermiller]

a = G * M
v(x) = √(2 *a)/x

This equation is incorrect. It should read:
$$v(x)=-\frac{dx}{dt}=\sqrt{2GM}\sqrt{\frac{1}{x}-\frac{1}{x_0}}=\sqrt{\frac{2GM}{x_0}}\sqrt{\frac{x_0-x}{x}}$$
where x₀ is the starting radius.

Chet

 

[rcgldr]

Go back to post #50 to see how to continue from here.

 

[Shahar]

Go back to post #50 to see how to continue from here.

Ok, I read post #50 a few times and I am lost. I don't understand how the equations contributor to the solution.

 

[rcgldr]

Ok, I read post #50 a few times and I am lost. I don't understand how the equations contributor to the solution.

From post #28, you have this complicated result:

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

By using the substitution from post #34:
$$- \ \sqrt{\frac{r_0}{2 G M}} \sqrt{\frac{r}{r_0-r}} dr = dt $$
$$ r = r_0 \ sin^2(θ) $$
$$\sqrt{\frac{r}{r_0-r}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0-r_0 \ sin^2(θ)}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0(1\ -\ sin^2(θ))}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0\ cos^2(θ)}} = tan(θ) $$
$$dr = 2\ r_0 \ sin(θ)\ cos(θ) dθ$$
$$\sqrt{\frac{r}{r_0-r}} dr = (tan(θ)) \ (2 \ r_0 \ sin(θ)\ cos(θ) dθ) = 2 \ r_0 \ sin^2(θ) dθ $$
You end up with this equation to integrate:
$$-\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ = dt $$
and you need to convert r = r₀ or r₁ to θ = θ₀ or θ₁
$$θ = {sin^{-1}\left ( \sqrt{\frac{r}{r_0}} \right )} $$

 

[Chestermiller]

rcgldr:
It looks like Shahar is not yet advanced to the point where he can complete this integration. Why don't you complete the analytic solution and then calculate the time to fall to the planet? It would be interesting to compare this with the value of 2000 sec that Shahar estimated.

Shahar: The solution to this problem can be obtained graphically, by finding the area under the curve of 1/v(r) vs r. Could you please plot a graph of 1/v(r) (units of sec/meter) as a function of r (meters) and show us what you get? If you could, that would be great.

Chet

 

[rcgldr]

$$dt = -\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \int_{θ_0}^{θ_1}2 \ sin^2(θ) dθ = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ \ -\ sin(θ)cos(θ) \right ]_{θ_0}^{θ_1}$$

 

[Shahar]

$$dt = -\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ$$
$$t = \int_{θ_0}^{θ_1} -\sqrt{\frac{r_0}{2 \ G \ M}} \ (2 \ r_0 \ sin^2(θ)) dθ = -\sqrt{\frac{r_0}{2 \ G \ M}} \left [ 2\ r_0\ (\frac{1}{2}\ (θ\ -\ sin(θ)cos(θ))) \right ]_{θ_0}^{θ_1}$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{θ_0}^{θ_1}$$
$$r_0 = {10}^{14}, r_1 = {10}^7 $$
$$θ_0 = \frac{π}{2}, θ_1 = {10}^{-7} $$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{π/2}^{{10}^{-7}}$$

r₀ is 2*10^7. I made a mistake when I misread the question.
And θ₁ is only 0.00031 not 10^-7

 

[rcgldr]

$$r_0 = 2e^7 \ , \ r_1 = 1e^7 $$
$$θ_0 = π/2 \ , \ θ_1 = π/4$$
$$t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ M}} \left [ θ\ -\ sin(θ)cos(θ) \right ]_{π/2}^{π/4}$$
This result now matches the result in post #73:
$$t = -2000 ((π/4 - 1/2) -\ (π/2 - 0)) \simeq 2570.8 \ seconds $$