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 73600 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. answer should be in km/s

Name changes and homework dumps are annoying. Please show work for assistance

First off all ,, I believe that you are referring that we are the same person, unfortunately we are not. second this website as I can see from its name is Homework HELP. so if you willing to help do so, or else please don't reply to me. Third, if I know how to solve it I wouldn't have post the question here.

First off, we can tell that you are posting from ONE COMPUTER in Jiddah, Saudi Arabia.

Second, there are no proxies indicated for this IP address, and it's not indicating that you are posting from a school or university.

Third, one computer almost always indicates one student. Fooling around by posting with different names indicates that you are trying to fool someone. That's too bad.

Fourth, if all you do is post your entire assignment, with no evidence of thinking on your part, nothing will happen since no one here will do your work for you. But if you are specific about what you don't understand about the assignment or exactly what help you need, someone might be able to assist you. Ask specific questions!

To find the speed of the asteroid at closest approach, we can use the concept of conservation of energy. At an infinite distance from Earth, we assume the potential energy of the asteroid is zero. As it approaches closer, its potential energy decreases while its kinetic energy increases.

The formula for gravitational potential energy is given by:

PE = -GMm / r

where PE is the potential energy, G is the gravitational constant, M is the mass of the Earth, m is the mass of the asteroid, and r is the distance between the Earth's center and the asteroid.

We can assume the mass of the Earth (M) is constant and neglect the mass of the asteroid (m) compared to the Earth's mass. Thus, we can simplify the formula to:

PE = -GM / r

The total energy (E) of the asteroid is the sum of its kinetic energy (KE) and potential energy (PE):

E = KE + PE

At closest approach, the asteroid's kinetic energy is at its maximum while its potential energy is at its minimum (zero). Therefore, we have:

E = KE

Since the kinetic energy (KE) is given by:

KE = (1/2)mv^2

where v is the velocity of the asteroid, we can substitute this into the equation for total energy:

E = (1/2)mv^2

Now, let's calculate the speed in km/s. The gravitational constant G is approximately 6.674 × 10^-11 m^3/kg/s^2, and the mass of the Earth M is approximately 5.972 × 10^24 kg.

Step 1: Convert the distance from miles to kilometers.
73600 miles = 73600 * 1.60934 kilometers (since 1 mile = 1.60934 kilometers)

Step 2: Convert the Earth-asteroid distance (r) to meters.
73600 * 1.60934 kilometers = 73600 * 1.60934 * 1000 meters (since 1 kilometer = 1000 meters)

Step 3: Substitute the values into the equation for total energy.
E = (1/2)mv^2

Step 4: Calculate the speed (v).

E = KE
(1/2)mv^2 = -GM / r

v^2 = (-2GM / r)
v = √(-2GM / r) (taking the positive square root since speed cannot be negative)

Step 5: Convert the speed from m/s to km/s.

That should give you the speed of the asteroid in km/s at its closest approach to Earth.