Suppose an asteroid of mass 6e20 kg is nearly at rest outside the solar system, far beyond Pluto. It falls toward the Sun and crashes into the Earth at the equator, coming in at an angle of 30 degrees to the vertical, against the direction of rotation of the Earth. It is so large that its motion is barely affected by the atmosphere.
Impact Speed: 43474.4226
Calculate in hours the change in the length of a day due to the impact: ????
I have been trying this equation and solving for T and have gotten wrong answers each time:
-sin(theta)*R_1*M_a + 2/5*M_e*R_1^2*(2pi/8.64e4) = 2/5M_e*R_1^2(2pi/T)
R_1: 6.4e6
M_e: 6e24
8.64e4 = how many seconds in an hour
yes sorry!!!!! its supposed to be -sin(theta)*R_1*V_impact*M_a + ………
ugh i cant believe i left that out
Also, I only have one submission left for my HW so if anybody can back the first answer up that would be fantastic
Posted in 2012
Tagged Asteroid, atmosphere, Calculate, change, crashes, Day, Earth, equator, impa, kg, Length, length of a day, Mass, pluto, rest, rotation of the earth, solar system, sun, system