# Two meteoroids are heading for earth. Their speeds as they cross the moon's orbit are 3.5 km/s.?

Two meteoroids are heading for earth. Their speeds as they cross the moon's orbit are 3.5 km/s.?

(a) The first meteoroid is heading straight for earth. What is its speed of impact?

(b) The second misses the earth by 6000 km. What is its speed at its closest point?

by

(a) 0.5m(v2^2 - v1^2) = GMm*[1/R - 1/d] where R is radius of earth; d is distance of moon, M is mass of earth and m mass of meteoroid and v1 is initial velocity and v2 that at impact

or v2^2 = 3500^2 + 2*(6.67*10^-11)*(6*10^24)*10^-6[1/6.37 - 1/384] = 0.1225*10^8 + 1.236*10^8 or

v2^2 = 1.36*10^8 or v2 = 1.16*10^4 m/s = 11.6 km/s

(b) 0.5m(v2^2 - v1^2) = GMm*[1/(R+6*10^6) - 1/d] where R is radius of earth; d is distance of moon, M is mass of earth and m mass of meteoroid and v1 is initial velocity and v2 that at impact

or v2^2 = 3500^2 + 2*(6.67*10^-11)*(6*10^24)*10^-6[1/12.38 - 1/384] = 0.1225*10^8 + 0.626*10^8 or

v2^2 = 0.7482*10^8 or v2 = 0.865*10^4 m/s = 8.65 km/s

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by

You can find out how far away the moon's orbit is and what the earth's radius is then calculate the amount of potential energy that goes to increase the kinetic energy in each case : the problem is that you have to ignore the influence of the moon and the sun as there is no mention of where they are in relation to the earth and the meteoroids; the speed isn't terribly well defined either - I think that you will have to assume that it's relative to an earth centre fixed coordinate system and that there is no point in taking the earth's rotation into account or atmospheric drag when calculating the impact speed.

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+1 vote