Two meteoroids 250,000 km from Earth are moving at 2.1 km/s. Meteor one is headed in a straight path for Earth, while meteor two is on a path that will come within 8500 km from the Earth's center. (A) What is the speed of the first meteroid when it strikes Earth. (B) What is the speed of the second meteoroid at its closest approach to Earth. (c) Will the second meteroid ever return to Earth's vicinity?

I do not get this do you use something like v=sqr(GM/r)? I know part C is no, but how do you get part A and B. Any help would be greatly appreciated.

The sum of potential and kinetic energy is zero. Potential energy per mass (if zero at infinite distance) is -GM/R. Use the fact that the earth's radius is 6378 km. M is the mass of the Earth.

For meteor (1),
-GM/(250,000*10^3)+ (2100)^/2 =
-GM/6378*10^3)+ Vimpact^2/2

For meteor 2,
-GM/(250,000*10^3)+ (2100)^/2 =
-GM/8500*10^3)+ Vmax^2/2

Look up GM and solve.

If Vmax exceeds the escape velocity, it will not return, if this is considered a two-body problem. Because the sun plays a larger role than Earth at large meteor distances, the meteor could return to the same distance from the sun and the ecliptic plane of the Earth's orbit, but the chances are very small the Earth will be there at the time.

A: To calculate the speed of the first meteoroid when it strikes Earth, we can use the principle of conservation of energy. Since the meteoroid is moving in a straight path towards Earth, it means there is no change in its potential energy. Therefore, the sum of its initial kinetic energy and gravitational potential energy equals its final kinetic energy at Earth's surface.

The initial kinetic energy of the meteoroid is given by:
KE_initial = (1/2) * m * v_initial^2

The potential energy at a distance of 250,000 km from Earth's center is given by:
PE_initial = - G * (m * M) / r

Considering that Earth's radius (r) is approximately 6,400 km and the mass of Earth (M) is roughly 5.972 × 10^24 kg, we have:
PE_initial = - G * (m * M) / (r + 250,000 km)

Setting the initial kinetic energy equal to the final kinetic energy (which is the total energy when it strikes Earth's surface):
KE_initial = KE_final = (1/2) * m * v_final^2

Since there is no change in potential energy when it strikes Earth, the equation becomes:
KE_initial = (1/2) * m * v_initial^2 = KE_final = (1/2) * m * v_final^2

By simplifying:
v_final^2 = v_initial^2

Therefore, the speed of the first meteoroid when it strikes Earth (v_final) is equal to the speed it had initially (v_initial).

B: To determine the speed of the second meteoroid at its closest approach to Earth, we can use the same principle of conservation of energy. However, in this case, we need to consider the change in potential energy since the meteoroid's distance from Earth's center is changing.

The initial kinetic energy of the meteoroid is again given by:
KE_initial = (1/2) * m * v_initial^2

The potential energy at a distance of 250,000 km from Earth's center is given by:
PE_initial = - G * (m * M) / (r + 250,000 km)

To calculate the potential energy at the closest approach (8500 km from Earth's center), we have:
PE_final = - G * (m * M) / (r + 8500 km)

Setting the initial kinetic energy plus initial potential energy equal to the final kinetic energy plus final potential energy:
KE_initial + PE_initial = KE_final + PE_final

By rearranging and solving for v_final, we can find the speed of the second meteoroid at its closest approach.

C: As you mentioned, the answer to part C is no. The second meteoroid will not return to Earth's vicinity because its closest approach does not bring it within Earth's atmosphere or gravitational influence to cause it to be captured or pulled back. It will continue on its path away from Earth.

To solve this problem, we need to use the conservation of angular momentum for an object in orbit. The equation is given by:

m1v1r1 = m2v2r2

Where:
m1 and m2 are the masses of the two meteoroids (assumed to be the same),
v1 and v2 are their velocities,
and r1 and r2 are their respective distances from the center of the Earth.

(A) To find the speed of the first meteoroid when it strikes Earth, we can consider its distance from Earth as the radius of the Earth (RE = 6,371 km). Thus, r1 = RE. We know that r2 = 8,500 km and v2 = 2.1 km/s. However, we need to find v1, the velocity of the first meteoroid when it strikes Earth.

Using the conservation of angular momentum equation, we have:

m1v1r1 = m2v2r2

Since m1 = m2 and r1 = r2, we can simplify the equation to:

v1 = (v2 * r2) / r1

Plugging in the values, we have:

v1 = (2.1 km/s * 8,500 km) / 6,371 km
v1 ≈ 2.8 km/s

Therefore, the speed of the first meteoroid when it strikes Earth is approximately 2.8 km/s.

(B) To find the speed of the second meteoroid at its closest approach to Earth, we can consider its distance from Earth as the sum of the Earth's radius and r2, as it comes within 8,500 km of the Earth's center. Thus, r1 = RE + r2.

Using the conservation of angular momentum equation, we still have:

m1v1r1 = m2v2r2

Plugging in the values, we have:

m1v1 * (RE + r2) = m2v2r2

Now, we solve for v2:

v2 = (m1v1 * (RE + r2)) / (m2r2)

Since m1 = m2, we can simplify the equation to:

v2 = (v1 * (RE + r2)) / r2

Plugging in the values, we have:

v2 = (2.8 km/s * (6,371 km + 8,500 km)) / 8,500 km
v2 ≈ 6.1 km/s

Therefore, the speed of the second meteoroid at its closest approach to Earth is approximately 6.1 km/s.

(C) No, the second meteoroid will not return to Earth's vicinity because its trajectory does not intersect with the Earth's path. It will continue on its own path through space.

To solve this problem, you can indeed use the formula v = sqrt(GM/r) where v is the speed of the meteoroid, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of the Earth (approximately 5.972 × 10^24 kg), and r is the distance between the meteoroid and the center of the Earth.

(A) To find the speed of the first meteoroid when it strikes Earth, you need to calculate its speed when it is at a distance of 0 km from Earth's center. Plugging the values into the formula, you get:

v1 = sqrt(GM/r1)
= sqrt((6.67430 × 10^-11 m^3 kg^-1 s^-2) × (5.972 × 10^24 kg) / (250,000 km + 6,371 km))
= sqrt(4.021 × 10^14 m^2 s^-2)
≈ 6.34 km/s

Therefore, the speed of the first meteoroid when it strikes Earth is approximately 6.34 km/s.

(B) To find the speed of the second meteoroid at its closest approach to Earth, you need to calculate its speed when it is at a distance of 8,500 km from Earth's center. Plugging the values into the formula, you get:

v2 = sqrt(GM/r2)
= sqrt((6.67430 × 10^-11 m^3 kg^-1 s^-2) × (5.972 × 10^24 kg) / (8,500 km + 6,371 km))
= sqrt(5.596 × 10^12 m^2 s^-2)
≈ 2.36 km/s

Therefore, the speed of the second meteoroid at its closest approach to Earth is approximately 2.36 km/s.

(C) As you correctly mentioned, the second meteoroid will not return to Earth's vicinity because its closest distance from Earth's center is 8,500 km, which means it will not encounter any gravitational forces strong enough to pull it back towards Earth.

Please note that the formula used here assumes a point-mass approximation for Earth, ignoring its shape and any other gravitational influences.