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.
Yes, sort of
(1/2)m v^2 = original (1/2) m v^2 + loss of potential energy as the mass nears earth
That loss is
(G M m/r far away - G M m /r near)
for r far away use 2.5*10^8
for r near use radius of earth for the one that hits
for r near use radius of earth + 8.5*10^6 for the one that comes close
-(G M m/r far away - G M m /r near)
To find the speed of the first meteoroid when it strikes Earth (Part A), we can use the conservation of energy principle. The potential energy at the initial distance (250,000 km) is equal to the kinetic energy just before striking Earth.
Step 1: Find the initial gravitational potential energy:
Gravitational potential energy (U) can be calculated using the formula U = GMm/r, where G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of Earth (5.972 × 10^24 kg), m is the mass of the meteoroid (assuming it is negligible), and r is the initial distance between Earth and the meteoroid (250,000 km or 250,000,000 meters).
U = (6.67430 × 10^-11) * (5.972 × 10^24) * m / 250,000,000
Step 2: Find the kinetic energy just before striking Earth:
Kinetic energy (K) can be calculated using the formula K = 1/2 * m * v^2, where m is the mass of the meteoroid (assuming it is negligible), and v is the velocity of the meteoroid just before striking Earth.
K = 1/2 * m * (2.1)^2
Step 3: Equate potential energy and kinetic energy:
Since energy conservation tells us that potential energy is equal to kinetic energy, we can set U equal to K and solve for v.
(6.67430 × 10^-11) * (5.972 × 10^24) * m / 250,000,000 = 1/2 * m * (2.1)^2
Simplifying the equation, you can cancel out the mass (m) and solve for v.
v = sqrt((6.67430 × 10^-11) * (5.972 × 10^24) / 250,000,000 * (2.1)^2)
Solving this equation will give you the speed of the first meteoroid when it strikes Earth (Part A).
To find the speed of the second meteoroid at its closest approach to Earth (Part B), you can use the conservation of mechanical energy. At its closest approach, the potential energy is equal to the sum of kinetic energy and the potential energy at a distance of 8500 km from Earth's center.
Step 1: Find the final gravitational potential energy:
The potential energy at a distance of 8500 km from Earth's center is Uf = GMm/rf, where rf is the final distance between Earth's center and the meteoroid (8500 km or 8,500,000 meters).
Step 2: Find the initial gravitational potential energy:
The potential energy at the initial distance of 250,000 km (250,000,000 meters) is Ui = GMm/ri, where ri is the initial distance between Earth's center and the meteoroid.
Step 3: Find the initial kinetic energy:
Using the initial distance and velocity (2.1 km/s), you can find the initial kinetic energy (Ki).
Ki = 1/2 * m * (2.1)^2
Step 4: Equate the initial and final energies:
Using the conservation of mechanical energy, you can set (Ui + Ki) equal to Uf and solve for vf (the velocity at the closest approach).
(GMm/ri) + (1/2 * m * (2.1)^2) = GMm/rf
Canceling out the mass (m) and solving for vf, you will get the speed of the second meteoroid at its closest approach to Earth (Part B).
For Part C, the second meteoroid will not return to Earth's vicinity. This is because its closest approach (8,500 km) is greater than the radius of Earth (approximately 6,371 km). Therefore, it will not be able to overcome Earth's gravitational pull and return to Earth.