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: http://www.youtube.com/watch?v=JrM4KVNwB94

Youtube thumbnail

&feature=channel

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 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 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.

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.

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;_ylt=Amrjp...

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.

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_Avir...

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

7 Answers

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  • 10 years ago
    Best Answer

    Initial data and assumptions:

    M= 1.9891×10^30 kg

    G= 6.67428x10^-11 kg

    Ro = 1.49643*10^11 m

    vo = 29785.3 m/s

    We assume Earth's orbit is circle with radius Ro. Value for Ro is slightly adjusted here to match other data, but it's still accurate enough.

    ------------------------------

    Conservation of energy:

    -GMm/Ro + mvo^2/2 = -GMm/Rmax

    Because on circular orbit centrifugal force and gravity are balanced:

    mvo^2/Ro = GMm/Ro^2

    multiply by Ro/2

    mvo^2/2 = GMm/(2Ro)

    plug in equation for energy:

    -GMm/Ro + GMm/(2Ro) = -GMm/Rmax

    divide by -GMm:

    1/Ro - 1/(2Ro) = 1/Rmax

    Rmax = 2Ro

    Third Keppler's Law:

    P^2/a^3 = Po^2/Ro^3

    P = Po √(a^3/Ro^3) .......(*)

    When probe is moving directly away from the sun, we may consider its trajectory as limit case of a degenerate ellipse with an eccentricity very close to 1. With aphelion held at same distance while increasing eccentricity, the ellipse becomes a line of length R. The semi-major axis is half the width of the ellipse along the long axis, which in the degenerate case becomes Rmax/2=Ro

    when e-->1 then Rmin -->0, a-->Rmax/2=Ro,

    and from (*)

    P-->Po

    P means time for complete revolution, or in our case from Rmax to the Sun and back. Probe, falling from aphelion only completes half the orbit, hence time required for probe to fall into the Sun from distance Rmax is

    t1=P/2=365.25/2=182.625 days

    Notice that when probe intercepts Earth's orbit for the second time, on its way to the Sun, probe's speed will again be vo, but now in opposite (inward direction). This will be important later.

    Repeat same procedure assuming the probe is falling toward the Sun from Earth's orbit with zero initial speed:

    Rmin -->0, a-->Ro/2

    P --> Po √((Ro/2)^3/Ro^3)= Po √8 / 8

    time to reach the Sun would be:

    t2=P/2=Po√8/16= 64.5677 days=5578648 s

    average speed: vff_av=1.49643*10^11/ 5578648 = 26824.2 m/s

    Same as above, moving from the Earth directly toward the Sun but this time with initial speed vo:

    net average speed is:

    v_av = vo + vff_av = 29785.3 + 26824.2 = 56609.5 m/s

    time required for probe to fall into the sun is:

    t3 = Ro/v_av = 1.49643*10^11/ 56609.5 = 2643426 s = 30.595 days

    Now let's get back to case of outward motion. We know that from distance Rmax=2Ro it takes 182.625 days to fall into the Sun, and of that time t3=30.595 days is spent on section from Earth's orbit to the Sun. That means it takes

    t4 = t1 - t3 = 182.625 - 30.595 = 152.03 days

    for probe to reach Earth's orbit falling from aphelion (Rmax), and same time is required for probe to reach aphelion in the first place.

    Finally, the answer to your question is:

    T = t4 + t1 = 152.03 + 182.625 = 334.655 days

    I'll check the result later.

  • 10 years ago

    Maybe you got put in a corner because you don't know how to describe the problem so others know what the hell you're asking.

    First Grade Rocks added:

    "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."

    Alright; 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? The answer will be different depending on the case. And since you weren't able to make yourself clear the first time, I think it's prudent to make sure you don't have a typo in there.

    By the way, what exactly does it mean to be "tangential to the sun"? See what I mean? Your articulative skills need some fine-tuning.

    And, incidentally, the example of a projectile fired from Earth isn't quite the same thing. 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. It wouldn't just move purely radially away from or toward the Sun. It would enter an elliptical solar orbit.

  • 4 years ago

    O.K., for starters-if the Earth stopped rotating on its axis, there would be little movement in the atmosphere as well as the oceans. This would impact both our weather and the temperature of the atmosphere-plus its ability to remove impurities (smog, volcanic ash, etc.). If the Earth were to stop in its orbit around the Sun, it would not have enough forward velocity=centrifugal force, to keep a solid path. Eventually gravity (either from the Sun or outer space), would take over, at which point it's a one-way ticket to either of those. So far, I did not see any mention of the Moon in these answers. Since it is held in tension bet. the Earth and the Sun, it would most likely hit Earth in both cases. That's a bad hair day. Afterthought-it might even affect plate movement under the Earth's crust, too. Not that it would matter-since all people would be gone. This (plate tectonic factor), is another question...

  • 10 years ago

    I don't think you need to do the calculus. It has been done for you in the classic equations of motion. You could solve this several ways, perhaps easiest to determine the highest point reached using s = u^2/2g. Then use s = s0 +v0t + (1/2)gt^2 and solve for t.

    But you should probably start by using the correct initial velocity u = 2pi AU / year, not 1 AU / year.

    ADDED: Closing the question? The confusion was not mine. What probe?

    ADDED (2): You are right. I thought about that after I went to bed. So I guess you will still need to integrate over the distance to the Sun.

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  • 10 years ago

    i can't give you the answer but i want to enlighten the fact that the operation of changing the speed vector direction of the earth will highly affect its spin. the only way the earth can change its direction is by "rolling" on the quarter pipe.

    this would have many consequences, like tectonic stuff and everything.

  • Koshka
    Lv 4
    10 years ago

    Goodness!! Don't we have enough problems as is, without encountering a cosmic quaterpipe?

  • Anonymous
    10 years ago

    I agree with "Martin".

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