Problem using integral form of Work K.E. Thm

[skate_nerd]

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-.5mv_o²=∫[from x_o to x]F(x)dx.
Where m=the mass of the earth, v_o=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 x_o→x to r_AU→r(t). Solved the integral and got
v(x)=sqrt(-2GM((1/r_AU)-(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 r_AU and the final point I'm trying to get the Earth to is the radius of the sun, ill just write as r_o. The center of the sun is the origin of the coordinate system.

 

[ehild]

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-.5mv_o²=∫[from x_o to x]F(x)dx.
Where m=the mass of the earth, v_o=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 x_o→x to r_AU→r(t). Solved the integral and got
v(x)=sqrt(-2GM((1/r_AU)-(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 r_AU and the final point I'm trying to get the Earth to is the radius of the sun, ill just write as r_o. The center of the sun is the origin of the coordinate system.

Is r(t) smaller or larger than r_AU during the motion of the Earth towards the sun? So what is the sign of -(1/r_AU -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

 

[skate_nerd]

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/r_AU])).
I need to find t(r), which would be
t(r)=∫(from r_AU to r_sun)dr/v(r).
→=1/(sqrt(2GM))∫(from r_AU to r_sun)dr/(sqrt([1/r(t)]-[1/r_AU])
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.

 

[ehild]

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

 

[skate_nerd]

Maybe I'm missing something, but putting a negative sign before all of that wouldn't help me avoid getting an imaginary answer.

 

[ehild]

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

 

[skate_nerd]

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?

 

[ehild]

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