# If earth suddenly encountered a cosmic quarter-pipe so its velocity was now radially outward, how (cont))?

If earth suddenly encountered a cosmic quarter-pipe so its velocity was now radially outward, how long would it take to fall into the sun?
I've solved it using calculus, but we aren't suppose to know that stuff yet. Teacher made me sit in a corner last time I tried to use calculus.
quarter-pipe:...
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If earth suddenly encountered a cosmic quarter-pipe so its velocity was now radially outward, how long would it take to fall into the sun?

I've solved it using calculus, but we aren't suppose to know that stuff yet. Teacher made me sit in a corner last time I tried to use calculus.

quarter-pipe: http://www.youtube.com/watch?v=JrM4KVNwB...

I've solved it using calculus, but we aren't suppose to know that stuff yet. Teacher made me sit in a corner last time I tried to use calculus.

quarter-pipe: http://www.youtube.com/watch?v=JrM4KVNwB...

Update:
Ok, Martin,
Earth's orbital radius is 1au and its period is 1 year. Instead of its orbital velocity being tangential to the sun, the velocity is changed to radial.
Or if that is too big to wrap your head around, just imagine a probe being fired off of earth with the same speed as earth has in its orbit...
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Ok, Martin,

Earth's orbital radius is 1au and its period is 1 year. Instead of its orbital velocity being tangential to the sun, the velocity is changed to radial.

Or if that is too big to wrap your head around, just imagine a probe being fired off of earth with the same speed as earth has in its orbit around the sun, except the speed is now directed radially instead of tangential to the sun.

Earth's orbital radius is 1au and its period is 1 year. Instead of its orbital velocity being tangential to the sun, the velocity is changed to radial.

Or if that is too big to wrap your head around, just imagine a probe being fired off of earth with the same speed as earth has in its orbit around the sun, except the speed is now directed radially instead of tangential to the sun.

Update 2:
1. "but you still need to clarify yourself a bit. You say you want to redirect Earth's path so its velocity is outward, away from the Sun? Or do you really mean inward, toward the Sun?"
I said "outwards" and I meant outwards, but if you can figure it out for inwards, it is not that...
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1. "but you still need to clarify yourself a bit. You say you want to redirect Earth's path so its velocity is outward, away from the Sun? Or do you really mean inward, toward the Sun?"

I said "outwards" and I meant outwards, but if you can figure it out for inwards, it is not that difficult to then solve for outwards.

2. "By the way, what exactly does it mean to be "tangential to the sun"? See what I mean?"

I dropped into polar coords. Tangential is along the path of earth's orbit. Radial is directed away from the focus of the orbit -- which is the sun. Instead of the probe following earth's orbit. (which would be tangential), it is now radial. If you assume that earth has an orbit close to circular, than tangential becomes perpendicular to radial.

3. "Even with an initial velocity exactly parallel to the Earth-Sun radius as reckoned relative to the Earth-local frame the projectile would still have orbital angular momentum which must be conserved."

I said "outwards" and I meant outwards, but if you can figure it out for inwards, it is not that difficult to then solve for outwards.

2. "By the way, what exactly does it mean to be "tangential to the sun"? See what I mean?"

I dropped into polar coords. Tangential is along the path of earth's orbit. Radial is directed away from the focus of the orbit -- which is the sun. Instead of the probe following earth's orbit. (which would be tangential), it is now radial. If you assume that earth has an orbit close to circular, than tangential becomes perpendicular to radial.

3. "Even with an initial velocity exactly parallel to the Earth-Sun radius as reckoned relative to the Earth-local frame the projectile would still have orbital angular momentum which must be conserved."

Update 3:
That is the reason we are in a sun-centric coordinate system. Radial means radial to the sun, not the earth.

This is why I said it was "the speed is now directed radially instead of tangential to the sun." It is also why I chose the analogy of a quarter pipe to change the velocity be 90 degrees.

This is why I said it was "the speed is now directed radially instead of tangential to the sun." It is also why I chose the analogy of a quarter pipe to change the velocity be 90 degrees.

Update 4:
To simplify everything. The probe is now moving directly away from the sun (in the sun's frame of reference) and it will eventually fall directly back into the sun.

The initial magnitude of the velocity is

v = 1au/year and the probe is 1 au from the sun.

The initial magnitude of the velocity is

v = 1au/year and the probe is 1 au from the sun.

Update 5:
******************

I will be closing this question shortly in that it has grown too confusing. I have reasked it here if anyone wishes to answer: http://answers.yahoo.com/question/index;...

I will be closing this question shortly in that it has grown too confusing. I have reasked it here if anyone wishes to answer: http://answers.yahoo.com/question/index;...

Update 6:
**********

Thank you Larry,

You are correct, I intended to have it moving at 2π au/yr ... but messed up. However, you are using an equation with constant g. Here we need to use a = GM/r^2. But you can use Kepler's laws to solve this.

Thank you Larry,

You are correct, I intended to have it moving at 2π au/yr ... but messed up. However, you are using an equation with constant g. Here we need to use a = GM/r^2. But you can use Kepler's laws to solve this.

Update 7:
The answer is:
=(π/2 + π/4 +1/2) / π *1yr
= 0.90915 yr
= 332 days
You take a degenerate ellipse and expand the minor axis like this:
http://i278.photobucket.com/albums/kk114/Remo_Aviron/Kepler.jpg
This dufus shows how calculus and kepler come up with the same answer to a very similar problem....
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The answer is:

=(π/2 + π/4 +1/2) / π *1yr

= 0.90915 yr

= 332 days

You take a degenerate ellipse and expand the minor axis like this:

http://i278.photobucket.com/albums/kk114...

This dufus shows how calculus and kepler come up with the same answer to a very similar problem. http://i278.photobucket.com/albums/kk114...

=(π/2 + π/4 +1/2) / π *1yr

= 0.90915 yr

= 332 days

You take a degenerate ellipse and expand the minor axis like this:

http://i278.photobucket.com/albums/kk114...

This dufus shows how calculus and kepler come up with the same answer to a very similar problem. http://i278.photobucket.com/albums/kk114...

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