The gravitational pull of the earth on an object is inversely proportional to th
ID: 1364896 • Letter: T
Question
The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight mg , where g=9.8m/s2, and at large distances, the force is zero.
Part A
If a 80000-kg asteroid falls to earth from a very great distance away, how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.
Express your answer using two significant figures.
art B
What will be its minimum speed as it strikes the earth's surface?
Express your answer using two significant figures.
The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight mg , where g=9.8m/s2, and at large distances, the force is zero.
Part A
If a 80000-kg asteroid falls to earth from a very great distance away, how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.
Express your answer using two significant figures.
art B
What will be its minimum speed as it strikes the earth's surface?
Express your answer using two significant figures.
Explanation / Answer
(A)
The general form of the gravitational potential energy of mass m = -GMm/r
Where,
M is Mass of Earth.
m is mass of asteroid = 80000 Kg
r = Radius of Earth
Kinetic Energy it will Imaprt to the surface of Earth , K = GMm/r
K = 6.674 * 10^-11 * 5.972 * 10^24 * 80000/ (6371 * 10^3)
KINETIC ENERGY = 5.0 * 10^12 J
(B)
Minimum speed = 0.5 * m*v^2 = 5.0 * 10^12
Minimum Speed , v = sqrt((5.0 * 10^12 *2) / (80000)) m/s
Minimum Speed , v = 11180 m/s
Minimum Speed , v = 1.12 * 10^4 m/s
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