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.

(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.)

F = G m Me/r^2

potential energy relative to zero at infinity = -G m Me/r
here r = 84600 miles
convert that to meters
then
potential energy lost = Kinetic energy gained
G m Me/r = (1/2)m v^2
so
v^2 = 2 G Me/r

for part b you need m
diameter = 2*10^3
so
diameter/2 = radius = 1,000 meters
convert density to kg/m^3
3.33 g/cm^3 * (1kg/10^3g) * (10^6 cm^3/m^3) = 3330 kg/m^3
m = density * (4/3)pi (diameter/2)^3

then do Ke = (1/2) m v^2

This is where I posted the reply to the multiple postings.

To find the speed of the asteroid at closest approach, we can use the law of conservation of energy. We know that the asteroid's speed at infinite distance is zero, so we can consider only its interaction with the Earth. At closest approach, the kinetic energy of the asteroid is equal to the potential energy at that distance.

(a) To find the speed, we need to find the potential energy at that distance and equate it to the kinetic energy.

The potential energy of an object near the surface of the Earth is given by the equation:

PE = -GMm/R

Where PE is the potential energy, G is the gravitational constant (approximately 6.67 x 10^-11 Nm^2/kg^2), M is the mass of the Earth (approximately 5.97 x 10^24 kg), m is the mass of the asteroid, and R is the distance between the center of the Earth and the asteroid.

At closest approach, R = 84600 miles = 135996.5 km = 1.359965 x 10^8 m (converting miles to meters).

The mass of the asteroid is not given in the question, so we will need to assume a value. Let's assume a mass of 1 kg for simplicity.

Substituting the values into the equation, we have:

PE = - (6.67 x 10^-11 Nm^2/kg^2) * (5.97 x 10^24 kg) * (1 kg) / (1.359965 x 10^8 m)

Simplifying the equation, we have:

PE = - (6.67 x 10^-11 Nm^2/kg^2) * (5.97 x 10^24 kg) * (1 kg) / (1.359965 x 10^8 m)
≈ -4.899 x 10^6 J

At closest approach, the potential energy is equal to the kinetic energy, so:

KE = -4.899 x 10^6 J

The kinetic energy of an object is given by the equation:

KE = 0.5mv^2

Where KE is the kinetic energy, m is the mass of the object, and v is its velocity.

Substituting the values, we have:

0.5 * (1 kg) * v^2 = -4.899 x 10^6 J

Solving for v, we get:

v^2 = (-4.899 x 10^6 J) / (0.5 * 1 kg)
v^2 ≈ -9.798 x 10^6 m^2/s^2

Since velocity cannot be negative, we take the positive square root:

v ≈ 3129 m/s

Therefore, the speed of the asteroid at closest approach is approximately 3129 m/s.

(b) To estimate the kinetic energy of the asteroid at closest approach, we can use the equation:

KE = 0.5mv^2

Given that the diameter of the asteroid is 2.0 km, the radius can be calculated as 1.0 km or 1000 m.

The mass of the asteroid can be calculated using the volume of a sphere:

V = (4/3)πr^3

V = (4/3)π(1000 m)^3 = (4/3)π(1 x 10^9 m^3)

Assuming the average density of the asteroid is 3.33 g/cm^3, we can convert this to kg/m^3:

density = 3.33 g/cm^3 = 3.33 x 10^3 kg/m^3

Using the density and volume, we can find the mass:

mass = density x volume = (3.33 x 10^3 kg/m^3) x (4/3)π(1 x 10^9 m^3)

Simplifying the equation, we have:

mass ≈ 1.396 x 10^12 kg

Substituting the mass and speed into the equation for kinetic energy:

KE = 0.5 x (1.396 x 10^12 kg) x (3129 m/s)^2

Calculating the equation, we get:

KE ≈ 6.556 x 10^18 J

Therefore, the estimated kinetic energy of the asteroid at closest approach is approximately 6.556 x 10^18 J.