In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after it had passed the Earth. Suppose that at its closest approach, the asteroid was 84600 miles from the center of the Earth -- about a third of the distance to the Moon.
(a) Find the speed of the asteroid at closest approach, assuming its speed at infinite distance to be zero and considering only its interaction with the Earth.
? km/s
(b) Observations indicate the asteroid to have a diameter of about 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it has an average density of 3.33 g/cm3. (For comparison, a 1-megaton nuclear weapon releases about 5.6 1015 J of energy.)
? J
Check answer to Janie above
(a) To find the speed of the asteroid at closest approach, we can use the principle of conservation of energy. At its furthest distance from the Earth (assumed to be infinite distance), the asteroid has zero speed and potential energy. At closest approach, the potential energy is also zero. Therefore, the change in potential energy is zero and is converted entirely into kinetic energy.
We can find the speed of the asteroid using the equation for kinetic energy:
KE = 1/2 * mv^2
where KE is the kinetic energy, m is the mass of the asteroid, and v is the speed of the asteroid.
To find the mass of the asteroid, we can use the density of the asteroid and its diameter. The formula for the volume of a sphere is:
V = (4/3) * π * r^3
where V is the volume and r is the radius of the sphere. Since the diameter is given as 2.0 km, the radius is 1.0 km.
Now, we can calculate the mass of the asteroid:
V = (4/3) * π * (1.0 km)^3
V = 4/3 * π * 1.0 km^3
V = (4/3) * 3.14 * (1.0 km)^3
V = 4.19 km^3
To convert the mass to grams, we multiply by the density:
Mass = Volume * Density
Mass = 4.19 km^3 * (3.33 g/cm^3)
Mass = (4.19 km^3 * 10^6 cm^3/km^3) * (3.33 g/cm^3)
Mass = 13.91 x 10^6 g
Now, let's find the speed. Since the potential energy is converted entirely into kinetic energy, we can equate the two:
KE = 1/2 * mv^2
0 = 1/2 * (13.91 x 10^6 g) * v^2
Solving for v^2:
v^2 = 0 / (1/2 * 13.91 x 10^6 g)
v^2 = 0
Therefore, the speed of the asteroid at closest approach is zero km/s.
(b) To estimate the kinetic energy of the asteroid at closest approach, we can use the equation for kinetic energy:
KE = 1/2 * mv^2
We already calculated the mass of the asteroid to be 13.91 x 10^6 g. Since the speed of the asteroid at closest approach is zero (as calculated in part a), the kinetic energy is also zero.
Therefore, the kinetic energy of the asteroid at closest approach is zero J.