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Normally, it would be a parabola. But I was thinking what it would be if were to assume that Earth (and its atmosphere) offer no resistance to the stone's fall.

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    $\begingroup$ The effective earth mass of the gravitational force of the earth will diminish as the stone wades in, and the trajectory is trivially inferred from 285689. Plot it. $\endgroup$ – Cosmas Zachos Feb 21 '17 at 17:18
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The parabolic path that we get for projectile motions of rocks and other things is an approximation assuming that the gravitational field is of constant strength and direction.

If we truly considered the gravitational force as $\dfrac{-Gm_rM_E}{r^2}\hat{r}$, where $r$ is the distance from the rock mass, $m_r$, to the center of a spherically-distributed earth mass, $M_E$, ($\hat{r}$ points from the earth toward the rock at every instant), the path of the rock would depend on the total mechanical energy of the rock/gravitational system, $$E=\frac{1}{2} m_rv_r^2-\dfrac{Gm_rM_E}{r}.$$

  • If E<0, the path will be an ellipse.
  • If E=0, the path will be a true parabola.
  • If E>0, the path will be a hyperbola.

The question takes on a large difficulty if you actually let the rock fall through the mass distribution of the earth. This makes the mass term, $M_E$ in the gravitational force a function of $r$ (because the mass "above" the rock no longer affects the rock in this spherically symmetric problem). That additional $r$ behavior of mass causes the force to behave like linear restoring force rather than an inverse-square force. That will change the orbit quit a bit and I don't have time to work of the mathematics of the orbits. Maybe someone else can provide the details in a follow-up answer.

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  • $\begingroup$ May be we can consider Earth shrinking to an extremely dense point mass at its center, and then the stone was launched from a point on its old surface. Does that help? $\endgroup$ – cmb6 Feb 21 '17 at 16:52
  • $\begingroup$ In that case, my last paragraph is unnecessary. The first part doesn't change. $\endgroup$ – Bill N Feb 21 '17 at 16:54
  • $\begingroup$ On the second thought, may be we don't need to make such assumption. $\endgroup$ – cmb6 Feb 21 '17 at 16:55
  • $\begingroup$ The parametric equation for the trajectory inside the earth is not hard. Tangential velocity follows conservation of angular momentum; radial velocity increases in accordance with increased energy (log potential) $\endgroup$ – Floris Feb 23 '17 at 21:43
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Yes, its orbit would be elliptical, unless you project it in tangential direction(tangent to radius vector) with the orbit velocity at that point(its almost impossible to do so as there would be oscillations and it wont be exact circle) . In general it would be an elliptical orbit.
However you can make it parabolic by projecting it with escape velocity(should not be directed vertically upwards). Also it can be made hyperbolic if you throw it with speed greater than escape velocity.
I have ignored air resistance and gravitational force from other bodies. In basic kinematics we claim its path to be a parabola as normal projectiles rise to a very small height.

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    $\begingroup$ No it would not be elliptical, because (as @Bill N points out) inside the earth, the inverse square law no longer applies. $\endgroup$ – TonyK Feb 21 '17 at 17:04
  • $\begingroup$ Yea well I ignored that $\endgroup$ – Red Floyd Feb 22 '17 at 0:46
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Yes.

The stone would move along a very-very long and narrow ellipse. Center of where-previously-was-Earth would be at the far end of this ellipse.

UPDATE: this is in case if all the mass is concentrated in the Earth center.

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The stone will continue to move until the velocity of stone is opposed by air drag . If we assume air drag to be absent then stone will reach out of Earth atmosphere .And swirl around in the universe

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  • $\begingroup$ This doesn't answer the question about the shape of the orbit. $\endgroup$ – Bill N Feb 21 '17 at 16:23
  • $\begingroup$ why not an elliptic orbit that projects out of Earth on both sides by equal amount? $\endgroup$ – cmb6 Feb 21 '17 at 16:27
  • $\begingroup$ So the orbit will not have the elliptical path because the distance between the Earth and stone Will. Be large as compared to mass of the stone according to $F$=$G$$m$$M$/$(R+h)$$^2$.here(M= mass of Earth and $m$=mass of stone while R= radius of Earth and h= height covered by stone while G=gravitational constant) $\endgroup$ – Rishi Kakkar Feb 21 '17 at 16:42
  • $\begingroup$ Troller Rishi.. $\endgroup$ – Red Floyd Feb 21 '17 at 16:43

protected by Qmechanic Feb 21 '17 at 18:52

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