An asteroid is located 600,000 km away from Earth\'s surface, heading directly t

Question

An asteroid is located 600,000 km away from Earth's surface, heading directly toward the Earth at a speed of 5 km/s. What is the speed of the asteroid as it hits the Earth's surface? Assume that the Earth's atmosphere does not slow down the asteroid, and assume that the Earth is a perfect sphere. Do not plug values in until the very end. Use the following values:

ME = mass of the Earth = 5.97 x 1024 kg

RE = radius of the Earth = 6370 km.

G = Gravitational Constant = 6.673 x 10-11 m3 kg-1 s-2

The answer I get is: 12 km/s

At what minimum initial velocity would the payload from the problem above never come back to the Moon (i.e., its “escape velocity”)? Hint: The escape velocity is the minimum initial velocity which will allow the payload to ultimately reach r = . Once it reaches r = , its velocity would have been decreased to zero. Remember, at r = the gravitational potential is 0. What initial kinetic energy is required to achieve this? You can tell if you got the right answer by looking it up on Wikipedia – search for escape velocity of the moon.

Answer hint: This problem relies upon the conservation of energy. Potential and kinetic energy remain constant.

Explanation / Answer

Using law of conservation of energy

Energy at 600000 km = energy at the earth's surface

(-G*mE*m/r1) + (0.5*m*u^2) = (-G*mE*m/r2)+ (0.5*m*V^2)

m is the mass of the asteroid

m cancels

and r1 = RE + 600000 = 6380+600000 = 606380 km

r2 = 6380 km

G is the universal gravitational constant

then

(-(6.67*10^-11*5.97*10^24)/(606380*10^3)) + (0.5*5000^2) = (-6.67*10^-11*5.97*10^24/(6380*1000)) + (0.5*V^2)

V = 12.186 km/sec

to find the escape speed on the moon

Using law of conservation of energy

0.5*m*v^2= G*M*m/r

m is the mass of the body and M is the mass of the moon

m cancels

0.5*v^2 = G*M/r

v = sqrt(2Gm/r)

Escape speed is V = sqrt(2*G*M/r)

G is the universal gravitational constant

M is the mass of the moon = 7.34*10^22 Kg

r is the radius of the moon = 1.737*10^6 m

then V = sqrt(2*G*M/r) = sqrt((2*6.67*10^-11*7.34*10^22)/(1.737*10^6))

V = 2.374 km/sec

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