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? 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?
 A: 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:
$$ v \approx 43.5 \text{km/sec} $$
A: 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.
