Problem using integral form of Work-Energy Theorem

In summary, the conversation discusses the problem of the Earth losing all velocity and plummeting towards the sun. The solution involves using the Work K.E. Theorem and solving for the function v(r) to find the time it takes for the Earth to hit the sun. The conversation also includes a discussion on finding the time as a function of position using integrals. A possible solution is suggested using multiple substitutions to simplify the integral.
  • #1
skate_nerd
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The problem is that the Earth has lost all velocity and begins plummeting toward the sun. I need to find the time it takes for it to hit the sun.

Note: Primes indicate "dummy variables"

This solution begins with the Work K.E. Theorem:
$$\frac{1}{2}mv(x)^2-\frac{1}{2}mv_{o}^2=\int_{x_o}^{x}F(x')dx'$$
Where \(v_{o}=0\) and $$F(x')=F(r)=\frac{-GMm}{r^2}$$
Plugging it all in gives
$$\frac{1}{2}mv(r)^2-\frac{1}{2}m(0)^2=\int_{r_{au}}^{r(t)}-\frac{GMm}{r'^2}dr'$$
$$\frac{1}{2}mv(r)^2=-GMm\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'$$
Earth's mass cancels out, as expected, and then we want to solve to get the function \(v(r)\):
$$v(r)=\sqrt{-2GM\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'}$$
$$=\sqrt{-2GM(-\frac{1}{r(t)}+\frac{1}{r_{au}})}$$
$$=\sqrt{2GM(\frac{1}{r(t)}-\frac{1}{r_{au}})}$$
Next, we use the following formula to find time as a function of position:
$$t(r)=\int_{r_{au}}^{r_{sun}}\frac{1}{v(r')}dr'$$
$$=\frac{1}{\sqrt{2GM}}{\int_{r_{au}}^{r_{sun}}} \frac{1}{\sqrt{\frac{1}{r(t)}-\frac{1}{r_{au}}}}dr'$$
The above integral gets a very long and gross answer via wolframalpha, and when I try plugging in the bounds I end up with an imaginary number.
Anybody know where I went wrong?
 
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  • #2
Re: problem using integral form of Work K.E. Thm

skatenerd said:
$$\frac{1}{2}mv(r)^2-\frac{1}{2}m(0)^2=\int_{r_{au}}^{r(t)}-\frac{GMm}{r'^2}dr'$$
$$\frac{1}{2}mv(r)^2=-GMm\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'$$
Earth's mass cancels out, as expected, and then we want to solve to get the function \(v(r)\):
$$v(r)=\sqrt{-2GM\int_{r_{au}}^{r(t)}\frac{1}{r'^2}dr'}$$
$$=\sqrt{-2GM(-\frac{1}{r(t)}+\frac{1}{r_{au}})}$$
$$=\sqrt{2GM(\frac{1}{r(t)}-\frac{1}{r_{au}})}$$
Starting from here I would use v = dr/dt, giving
\(\displaystyle T = \int \frac{dr}{\sqrt{2GM \left ( \frac{1}{r} - \frac{1}{r_{au}} \right ) }} \)

There may be some substitutions that will shortcut this, but I prefer to work one substitution at a time. The method isn't that bad, but you really need to see it before you can do it on your own.

Let u = 1/r. Then the integral becomes
\(\displaystyle T = -\int \frac{1}{\sqrt{Au - B}}\frac{1}{u^2}du\)
(where A and B are the appropriate constants.)

Now let y = Au - B. Then
\(\displaystyle T = -A \int \frac{1}{\sqrt{y}} \frac{1}{(y + B)^2} dy\)

Let \(\displaystyle z = \sqrt{y}\). Then
\(\displaystyle T = -2A \int \frac{1}{(z^2 + B)^2} dz\)

Let \(\displaystyle z = \sqrt{B}~tan( \theta )\). Then
\(\displaystyle T = -\frac{2A}{B^{3/2}} \int cos^2( \theta )~d \theta \)

You can take it from here.

-Dan
 
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  • #3
Re: problem using integral form of Work K.E. Thm

Wow, yeah I don't think I would have ever come up with that on my own haha. Thanks for the help.
 
  • #4
Re: problem using integral form of Work K.E. Thm

skatenerd said:
Wow, yeah I don't think I would have ever come up with that on my own haha. Thanks for the help.
It highlights something I've noted in the more advanced Mathematics...keep redefining the problem until you get something familiar! (Cool)

-Dan
 
  • #5


One possible issue with this approach is that the Work-Energy Theorem assumes that the force is constant over the entire displacement. However, in this scenario, the force is changing as the Earth gets closer to the sun. This could lead to discrepancies in the calculations and potentially result in an incorrect answer.

Instead, it may be more appropriate to use the equations of motion, specifically the equation for gravitational force, to determine the time it takes for the Earth to hit the sun. This equation takes into account the changing force as the distance between the Earth and the sun decreases. Once the time is determined, it can be plugged into the integral form of the Work-Energy Theorem to calculate the work done and the change in kinetic energy.

It is also important to consider the assumptions made in this scenario, such as ignoring any external forces or the effects of the Earth's rotation and orbit around the sun. These may also affect the accuracy of the calculations.

Overall, it is important to carefully consider the assumptions and limitations of the problem and choose the appropriate equations and methods to accurately solve it.
 

FAQ: Problem using integral form of Work-Energy Theorem

What is the Work-Energy Theorem and how is it used?

The Work-Energy Theorem is a principle in physics that states the work done by all forces on an object is equal to the change in the object's kinetic energy. It can be used to solve problems involving the motion of an object and the forces acting on it.

How is the Work-Energy Theorem expressed in integral form?

The integral form of the Work-Energy Theorem states that the work done by a force on an object is equal to the change in the object's kinetic energy, which can be expressed as ∫Fdx = ΔKE. This means that the work done by the force is equal to the area under the force-displacement curve.

What are the limitations of using the integral form of the Work-Energy Theorem?

The integral form of the Work-Energy Theorem assumes that the forces acting on the object are constant, which may not always be the case in real-world situations. It also assumes that the object is moving in a straight line and that the forces are applied parallel to the displacement of the object.

Can the integral form of the Work-Energy Theorem be applied to systems with multiple objects?

Yes, the integral form of the Work-Energy Theorem can be applied to systems with multiple objects as long as the forces acting on each object are known and the motion of each object can be described separately. The total work done on the system will be equal to the change in the total kinetic energy of the system.

How does the integral form of the Work-Energy Theorem relate to conservation of energy?

The integral form of the Work-Energy Theorem is based on the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted into different forms. In this case, the work done on the object is equal to the change in its kinetic energy, which means that the total energy of the system remains constant.

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