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.)
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To find the speed of the asteroid at its closest approach, we can use the concept of conservation of energy. At infinite distance, the potential energy of the asteroid due to its interaction with the Earth is zero. Hence, at closest approach, all of the potential energy has been converted to kinetic energy.

(a) Finding the speed of the asteroid at closest approach:

1. First, we need to find the potential energy of the asteroid when it is at closest approach. The potential energy is given by the equation:

PE = -GMm/r

Where PE is the potential energy, G is the gravitational constant (approximately 6.674 × 10^(-11) Nm^2/kg^2), M is the mass of the Earth (approximately 5.972 × 10^24 kg), m is the mass of the asteroid (which we assume to be 1 kg for simplicity), and r is the distance between the center of the Earth and the asteroid.

Given that the distance is 84600 miles, we need to convert it to meters. 1 mile is approximately equal to 1609.34 meters. Therefore, the distance is:

r = 84600 miles * 1609.34 meters/mile

2. Next, we can calculate the potential energy:

PE = -GMm/r

3. Since the potential energy at infinite distance is zero, the change in potential energy is equal to the value of the potential energy at closest approach:

ΔPE = PE = -GMm/r

4. The change in potential energy is equal to the kinetic energy at closest approach:

ΔPE = KE = 0.5 * m * v^2

Where KE is the kinetic energy and v is the speed of the asteroid.

Setting the equations equal to each other:

0.5 * m * v^2 = -GMm/r

Cancelling out the mass (m) on both sides:

0.5 * v^2 = -GM/r

Multiplying both sides by 2:

v^2 = -2GM/r

Taking the square root of both sides:

v = √(-2GM/r)

Note: The negative sign in front of GM indicates that the potential energy is negative, implying attraction.

5. Plugging in the values for G, M, and r, we can calculate the speed (v) of the asteroid:

v = √(-2 * 6.674 × 10^(-11) Nm^2/kg^2 * 5.972 × 10^24 kg / r)

Remember that r should be in meters for consistent units.

(b) To estimate the kinetic energy of the asteroid at closest approach:

1. The formula for kinetic energy is:

KE = 0.5 * m * v^2

Where KE is the kinetic energy, m is the mass of the asteroid, and v is the speed of the asteroid.

2. The diameter of the asteroid is given as 2.0 km, which means the radius (r) is half of that.

r = 2.0 km / 2

3. Convert the diameter to meters:

r = (2.0 km / 2) * 1000 m/km

4. Since the density is given in grams per cubic centimeter, we need to convert the mass to kilograms. The volume (V) of the asteroid is given by:

V = (4/3) * π * r^3

1 cm^3 = 1 × 10^(-6) m^3 (conversion factor)

Convert the diameter to centimeters:

r_cm = (2.0 km / 2) * 100000 cm/km

1 g = 0.001 kg (conversion factor)

m = V * density

5. Calculate the kinetic energy:

KE = 0.5 * m * v^2

Where m is the mass in kg obtained in step 4, and v is the speed in m/s obtained in part (a).

This will give you the kinetic energy of the asteroid at closest approach.

Please note that the calculations involved are complex and require precise values for G, M, and r.

Check answer above.