Problem using integral form of Work K.E. Thm

In summary, the Earth loses all orbital velocity and falls towards the sun. To find the time it takes for the Earth to hit the sun, you need to solve an integral.
  • #1
skate_nerd
176
0

Homework Statement



I have this problem where the Earth immediately loses all orbital velocity and begins to fall towards the sun, and I need to find the time it takes for the Earth to hit it.
Seemed straight forward enough.

Homework Equations



Started with the work k.e. theorem,
.5mv(x)2-.5mvo2=∫[from xo to x]F(x)dx.
Where m=the mass of the earth, vo=0, and F(x)=F(r)=-GMm/r2 where M is the mass of the sun.

The Attempt at a Solution



So I made the bounds translate from xo→x to rAU→r(t). Solved the integral and got
v(x)=sqrt(-2GM((1/rAU)-(1/r(t))))
Seeing as how r(t) is never going to get bigger than 1 AU, this doesn't make any sense. The answer is already imaginary, and I haven't even gotten to the integral for solving for t(r) yet. Anybody know what I did wrong?

note: In case it wasn't that obvious, I'm using the initial position of the Earth as rAU and the final point I'm trying to get the Earth to is the radius of the sun, ill just write as ro. The center of the sun is the origin of the coordinate system.
 
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  • #2
skate_nerd said:

Homework Statement



I have this problem where the Earth immediately loses all orbital velocity and begins to fall towards the sun, and I need to find the time it takes for the Earth to hit it.
Seemed straight forward enough.

Homework Equations



Started with the work k.e. theorem,
.5mv(x)2-.5mvo2=∫[from xo to x]F(x)dx.
Where m=the mass of the earth, vo=0, and F(x)=F(r)=-GMm/r2 where M is the mass of the sun.

The Attempt at a Solution



So I made the bounds translate from xo→x to rAU→r(t). Solved the integral and got
v(x)=sqrt(-2GM((1/rAU)-(1/r(t))))
Seeing as how r(t) is never going to get bigger than 1 AU, this doesn't make any sense. The answer is already imaginary, and I haven't even gotten to the integral for solving for t(r) yet. Anybody know what I did wrong?

note: In case it wasn't that obvious, I'm using the initial position of the Earth as rAU and the final point I'm trying to get the Earth to is the radius of the sun, ill just write as ro. The center of the sun is the origin of the coordinate system.

Is r(t) smaller or larger than rAU during the motion of the Earth towards the sun? So what is the sign of -(1/rAU -1/r(t))?
You got the magnitude of the velocity from conservation of energy. How do you define the vector v in terms of dr/dt?

ehild
 
Last edited:
  • #3
Okay I looked at that wrong, however i still end up running into a problem later.
I eventually get that
v(r)=sqrt(2GM([1/r(t)]-[1/rAU])).
I need to find t(r), which would be
t(r)=∫(from rAU to rsun)dr/v(r).
→=1/(sqrt(2GM))∫(from rAU to rsun)dr/(sqrt([1/r(t)]-[1/rAU])
This is a very gross integral that I plugged into wolframalpha, and got a very long answer for, and when I plugged in the bounds I got imaginary numbers again.
 
  • #4
You miss a minus sign. dr/dt=-v=-sqrt(2GM([1/r(t)]-[1/rAU])), as r is decreasing.
You can find the same problem here in Physicsforums. Try to search. The integral can be done with suitable substitution.

ehild
 
  • #5
Maybe I'm missing something, but putting a negative sign before all of that wouldn't help me avoid getting an imaginary answer.
 
  • #6
You should not get imaginary answer. Using AU as unit of distance, and approximating the final distance as 0, type in int _1^0 (1/sqrt(1/x-1)dx) in Wolframalpha. As I see, Wolframalpha does integration by parts.

ehild
 
  • #7
Okay, that's somewhat satisfying. But how come if you don't approximate the final distance as zero, and instead make the bounds from 2 to 1 (since I was planning on going from 1 AU to the radius of the sun), you get an imaginary number?
 
  • #8
The bounds are from 1AU to the radius of the Sun, which is much less than 1 AU.
Use indefinite integral in wolframalpha and see what you get. The solution is real for x<1AU

I have to leave now.

ehild
 
Last edited:

1. What is the integral form of the Work Kinetic Energy Theorem?

The integral form of the Work Kinetic Energy Theorem is a mathematical expression that relates the work done by a net force on an object to its change in kinetic energy. It is represented as W = ∫F dx = ½mv2, where W is work, F is force, x is displacement, m is mass, and v is velocity.

2. How is the integral form of the Work Kinetic Energy Theorem derived?

The integral form of the Work Kinetic Energy Theorem is derived from the more familiar form of the theorem, which states that the net work done on an object is equal to the change in its kinetic energy. By integrating this equation with respect to displacement, we can express the work done as an integral of force over distance.

3. What are the limitations of using the integral form of the Work Kinetic Energy Theorem?

One limitation of the integral form of the Work Kinetic Energy Theorem is that it only applies to objects with constant mass and velocity. It also assumes that the net force and displacement are in the same direction. Additionally, it does not take into account other forms of energy such as potential energy.

4. In what situations is the integral form of the Work Kinetic Energy Theorem useful?

The integral form of the Work Kinetic Energy Theorem is useful in situations where the force acting on an object is not constant or where the object's mass and velocity are changing. It can also be used to calculate the work done by a varying force over a given distance.

5. How is the integral form of the Work Kinetic Energy Theorem used in real-world applications?

The integral form of the Work Kinetic Energy Theorem is used in real-world applications such as analyzing the motion of objects in physics and engineering. It is also used in industries such as transportation and construction to calculate the work done by forces on moving objects and structures. In addition, it is used in sports science to analyze the performance of athletes and their use of energy during physical activities.

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