What will be the velocity of a comet is falling to the Earth from infinity at the time of impact if Earth had no atmosphere? the comet is falling radially towards the earth.

Will this velocity be different for comets with different masses or same?

  • $\begingroup$ Is the comet falling radially toward the Earth? $\endgroup$ – joshphysics Feb 18 '14 at 18:40
  • $\begingroup$ Yes it is falling Radially $\endgroup$ – Deepak Nath Feb 18 '14 at 18:43
  • 2
    $\begingroup$ @joshphysics is there any other possibility starting from infinity? $\endgroup$ – DarioP Feb 18 '14 at 19:08
  • $\begingroup$ @DarioP I don't see why not. For example, consider firing the comet from a radial distance $r$ away with some miniscule velocity in the tangential direction. In this case, it's not immediately clear what the condition is for a collision with the Earth, but I would be surprised if such a collision couldn't be arranged by simply making this tangentially velocity sufficiently small for a given radius. Now compute the collision velocity and take $r\to \infty$. $\endgroup$ – joshphysics Feb 18 '14 at 19:13
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    $\begingroup$ @joshphysics when you take $r\rightarrow \infty$ even an infinitesimal tangential velocity will deviate you from falling into Earth. Earth just looks like a (mathematical) point from there, with just a tiny deviation you miss it and come back to infinity. Think also about the reversed motion: how could you reach infinity with some tangential velocity starting from the Earth surface? $\endgroup$ – DarioP Feb 18 '14 at 19:35

The two existing answers have both done the correct calculation, but both have forgotten to account for the Sun's gravity. A comet falling from the fringes of the Solar System is accelerated mainly by the Sun's gravity. We can see this from the expression for the potential energy at a distance $r$:

$$ V = -\frac{GMm}{r} $$

For the Sun $M = 1.9891 \times 10^{30}$ kilograms and $r$ (the orbital radius of the Earth) $\approx 1.5 \times 10^{11}$ so $V \approx 8.85 \times 10^8 m$ J.

For the Earth $M = 5.97219 \times 10^{24}$ kilograms and $r$ (the radius of the Earth) $\approx 6.4 \times 10^{6}$ so $V \approx 6.15 \times 10^7 m$ J.

So the effect of the Sun's gravity is about 14 times greater. The velocity of the comet will be given by:

$$ \frac{1}{2} mv^2 = (8.85 \times 10^8 + 6.15 \times 10^7)m $$

which gives:

$$ v \approx 43.5 \text{km/sec} $$

  • $\begingroup$ But that's not taking into account Earth's velocity about the sun. The comet could hit it head-on, or from the back, forming the bounds of the possible velocities. It can be anywhere between 12.4 and 72 km/s. For your number, it wouldn't even be at a right angle. $\endgroup$ – Alan Rominger Feb 18 '14 at 20:42
  • $\begingroup$ @AlanSE: true - This is all getting very complicated :-) $\endgroup$ – John Rennie Feb 19 '14 at 6:42
  • $\begingroup$ The computation is ok for the Earth-Sun escape velocity. However the probability that the comet will hit the Earth in an Earth-Sun system tends to zero as the initial distance goes to infinity, it IS zero if the comet is not on the Earth orbit plane. It will pass somewhere in between going close to the Sun (as real comets do). $\endgroup$ – DarioP Feb 19 '14 at 11:42

The gravitational energy of the comet at infinity gets converted into kinetic energy of the comet. Calling $m$ the mass of the comet, $M$ the mass of the Earth, $r$ the radius of the Earth we have:

$$G\frac{mM}{r} = \frac{1}{2}m v^2 $$

where $G$ is the gravitation constant and $v$ is the speed of the comet when it hits the surface. Thus:

$$ v = \sqrt{\frac{2GM}{r}} \approx 11\;\text{km/s}$$

This is also called escape velocity since if you can impress this velocity to an object, it will be able to escape from the Earth sphere of influence. As you can see it does not depend on the mass of the object.


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