A new summer disaster movie tells the story of an astronomer who notices an aste

Question

A new summer disaster movie tells the story of an astronomer who notices an asteroid passing through a point in space, 19 Earth radii away from the surface of the Earth, with a velocity 100 m/s perpendicular to a line from the asteroid to the center of the earth. The astronomer figures that the asteroid has been captured by the Earth’s gravitational field, and so it is now on an elliptic orbit around our planet. In order to see if the asteroid poses a real threat to the planet, the astronomer does a quick calculation. What is the distance of closest approach for the asteroid, and what is its velocity when it is at that distance? Is it time to panic?

Explanation / Answer

the total energy at 19R = total energy in orbit

in orbit

vo =sqrt(GM/r)

KE = 0.5*m*Vo^2 = KE = 0.5*m*G*M/r

PE = -G*M*m/r

total energy in orbit = -G*M*m/r

0.5*m*v^2 - G*M*m/19R= -G*M*m/r

0.5*100^2 - (6.67*10^-11*5.97*10^24/(19*6.37*10^6)) = -6.67*10^-11*5.97*10^24/(2r)

r = 6.1*10^7 m

(b)

orbital velocity = vo = sqrt(GM/r) = sqrt((6.67*10^-11*5.97*10^24/(6.1*10^7) = 2.55 km/s

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