Is this point at Hill sphere?

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Imagine that we sent some ball by cannon from the Earth directly to the Sun. To reach the sun the ball should have Escape Velocity-11,2 km/sec and due to Earth’s gravity this velocity will continuously decrease until the point where the gravity of the Sun and Earth are equal. After this point the ball will simply fall on the Sun. Could you please tell me at what distance is this point located from the Earth? What do we call this point? This point and all other points nearer (relative the Earth) are inside some imagery sphere, what do we call this sphere? Is it Hill sphere? :uhh:
 

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  • #2
Ich
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The point is 1.5 million kilometers away, called L1, and lies on the surface of the Hill sphere.
But there is on problem: All these things are defined in a rotating coordinate system. A cannon ball fired at the sun will actually never reach L1 due to the Coriolis force.
IOW, the speed of the cannon ball relative to the sun is not 11.2 km/s, there is also some 30 km/s earth orbital velocity to take into account, which changes the picture.
 
  • #3
D H
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The point is 1.5 million kilometers away, called L1, and lies on the surface of the Hill sphere.
No, and no.

The L1 point is outside of the Hill sphere, and the point where gravitational force from the Sun and Earth are equal is easily calculable but is of very little interest.
 
  • #4
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Ich
But there is one problem: All these things are defined in a rotating coordinate system. A cannon ball fired at the sun will actually never reach L1 due to the Coriolis force.
Well, I do not need the ball to reach the L1 point :) I need to find the point where moving ball (towards Sun) will be attracted by Sun’s gravity, of course if we fire the ball from cannon with some speed this speed will be decreased due to Earth’s (or Moon’s-if we fire ball from cannon located on the Moon) gravity and the ball will reach this point with much less speed, so what do we call this point and where is it located?

IOW, the speed of the cannon ball relative to the sun is not 11.2 km/s, there is also some 30 km/s earth orbital velocity to take into account, which changes the picture.
Yes, I know this :) but I am not interested in ball’s speed, I need to find the point where Sun’s gravity prevails over Earth’s one

D H
the point where gravitational force from the Sun and Earth are equal is easily calculable but is of very little interest.
So, where is this point? :)
is of very little interest.
I understand why-as far as I know nobody tried to reach Sun directly from the Earth (or Moon), but imagine that this is necessary. More precisely, we have got this kind of situation:
http://img30.imageshack.us/img30/8581/4545j.jpg [Broken]
From the Moon we fire ball at speed of 2.38 km/s (Moon Escape Velocity) directly to the Sun, this speed will be continuously decreased because of Moon’s and Earth’s gravity but then at some point in space Sun’s gravity will prevail and ball will keep falling on the Sun, so where is this point?
 
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  • #5
Ich
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D H said:
The L1 point is outside of the Hill sphere
I don't claim expertise in these definitions, but my sources (e.g. http://books.google.com/books?id=a_...CDUQ6AEwAQ#v=onepage&q="hill sphere"&f=false") tell me that L1 lies on the Hill sphere.
I don't know exactly what your first "no" was aimed at. L1 seem to be some 1.5 million kilometers away from earth, so this part is correct.
If you mean that it's incorrect to say that L1 is the point where the gravity of the Sun and Earth are equal, I agree. But it's IMHO the next best thing to such a point that males sense in the OP's context. Because, as you rightly say: "the point where gravitational force from the Sun and Earth are equal is easily calculable but is of very little interest."
Eagle9 said:
I need to find the point where moving ball (towards Sun) will be attracted by Sun’s gravity
at some point in space Sun’s gravity will prevail and ball will keep falling on the Sun
This would be correct if the Earth were not orbiting the sun. If you take orbital velocity into account, no such cannon ball (except maybe very fast ones) will fall on the sun. They all will circle the sun in an elliptical orbit.

If you had an object with the same angular velocity as the earth, somewhere between earth and sun, and let go of it, it would start falling towards the sun if it's outside (i.e. nearer the sun) L1, towards earth if it's inside. But all those objects would orbit the sun, or earth, or fall on earth.
 
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  • #6
D H
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I need to find the point where Sun’s gravity prevails over Earth’s one
This is beginning to sound like homework, so at this point we need to treat this as homework.

What are the relevant equations, Eagle9? What work have you done to solve this problem?
 
  • #7
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D H
It is not my homework for school :smile: I just want to know the answer to this question



Ich
no such cannon ball (except maybe very fast ones) will fall on the sun
So, what is speed is necessary for this purpose?
 
  • #8
D H
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L1 lies on the Hill sphere.
I stand corrected. The Hill sphere is tautologically defined as the distance to the L1 point.

D H
It is not my homework for school :smile: I just want to know the answer to this question
Telling a student the answer to a homework question without requiring the student to do some work on the problem is not conducive to learning. Our stance at this site is to treat questions that look like homework as if they are homework because just telling you the answer will not help you understand the problem.

I can give you a hint: Calculate the gravitational acceleration of the Moon toward the Sun and toward the Earth. Which is bigger?
 
  • #9
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D H
Telling a student the answer to a homework question without requiring the student to do some work on the problem is not conducive to learning.
I am not student :smile:

I can give you a hint: Calculate the gravitational acceleration of the Moon toward the Sun and toward the Earth. Which is bigger?
Frankly saying I have got no idea :shy:
 
  • #10
D H
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What is Newton's law of gravitation?
 
  • #11
Ich
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So, what is speed is necessary for this purpose?
FYI: Some 6000 km/s. But better keep going with DH's short course.
 
  • #12
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D H
What is Newton's law of gravitation?
The result of falling the apple at Newton’s head :shy:
Of course I know what is Newton's law :smile:

Ich
Some 6000 km/s
Six thousand kilometers per second? :eek:
 
  • #13
D H
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If you know what Newton's law of gravitation is, use it to calculate the acceleration of the Moon toward the Sun due to the Sun's gravity and the acceleration of the Moon toward the Earth due to the Earth's gravity.
 

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