Question

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of 5 x 10^9 kg. It is approaching the Earth on a head-on course with a velocity of 660m/s relative to the Earth and is now 6.0 x 10^6 km away.

With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

round to two significant figures.

Answer 1

The sum impact velocity, not accounting for changes of velocity due to changing gravitational attraction = 30,450 m/s

How and why

Everything needs to be in SI units.

Given values for the asteroid:

m = 5.0E+9 kg
va = 660 m/s
da = 5.5E+9 m

Also need

M = Mass of Sun
G = Universal Gravitational Constant
r = Radius of Earth's orbit

Values:

M = 1.9891E+30 kg
G = 6.67428E-11 m^3/kg-s^2

Radius needs to be calculated as circular, so I average aphelion and perihelion, which is pretty fair because we have the second most circular orbit in the Solar System to begin with.

Aphelion = 152,097,701 km
Perihelion = 147,098,074 km

Average = 149,597,887,500 m

So our total list of values comes down to:

m = 5.0E+9 kg
va = 660 m/s
da = 5.5E+9 m
Ms = 1.9891E+30 kg
G = 6.67428E-11 m^3/kg-s^2
r = 149,597,887,500 m

First need to find the orbital velocity of Earth along its orbit, given by:

v = SQRT { [GMs] / r ]

Solve

v = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (1.9891E+30 kg) ] / (149,597,887,500 m) }
v = SQRT { [ 1.3276 m^3/s^2 ] / (149,597,887,500 m) }
v = SQRT { 887,433,009 m^2/s^2 }
v = 29,790 m/s

Which is very close to the given value of 29.783 km/s (29,783 m/s), in fact, only 0.02% off. Not bad.

So the easy answer is that the impact velocity will be:

Vt = Vearth + Vast
Vt = (29,790 m/s) + (660 m/s)
Vt = 30,450 m/s

The more complicated answer is that Earth will be accelerating the asteroid toward it due to gravitational "pull."

The gravitational acceleration exerted by the Earth on the asteroid at the maximum distance is given by:

a = [GMe] / r^2

In this case, Me is mass of the Earth: 5.9736E+24 kg

a = [ (6.67429E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (5.5E+9 m)^2
a = [ 4.654E+14 m^3/s^2 ] / (3.025E+19 m^2)
a = 1.539E-5 m/s^2

So not much pull at that distance. The asteroid's pull on Earth is so negligible that it can be ignored. The force of gravitational attraction between the two objects at maximum distance is given by Newton's Law of Universal Gravitation:

F = [GMeMa] / r^2

r must be the distance from the center of each object, not just the given surface-to-surface. We don't know the radius of the asteroid, but we do know the radius of earth, so just add that distance to the total distance to get r

r = (5.5E+9 m) + (6371000 m) = 5,506,371,000 m

Solve

F = [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) * (5.0E+9 kg) ] / (5,506,371,000 m)^2
F = [ 1.9935E+24 m^3-kg/s^2 ] / (3.032E+19 m^2)
F = 65,748 N