A meteorite of mass 1*10^4kg is approaching head-on a planet of mass 5*10^29kg and radius 5*10^4km. Assume that the meteorite is initially at a very large distance from the planet where it has a speed 3*10^2km/s. Take 6.67*10^-11.

Determine the speed of the meteorite (in m/s) just before it hits the surface of the planet. (The planet has no atmosphere, so we can neglect all friction before impact)

v=

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please tell the formuka

I know its the initial KE+initial PE=final KE... but how do we calculate the initial PE? GMm/R.... but what R do we sue ehre? Its beens tated that the meteorite is already really far away from the planet initially..???

Given that its really far away from the planet you can assume that PE = 0

To determine the speed of the meteorite just before it hits the surface of the planet, we can use the principle of conservation of mechanical energy.

1. First, we need to find the potential energy at infinity (initial position) and at the surface of the planet (final position). The potential energy is given by:

Potential energy at infinity (initial position) = 0
Potential energy at the planet's surface (final position) = - GMm / R

Where G is the gravitational constant (6.67 x 10^-11 m^3/(kg*s^2)), M is the mass of the planet (5 x 10^29 kg), m is the mass of the meteorite (1 x 10^4 kg), and R is the radius of the planet (5 x 10^4 km or 5 x 10^7 m).

Potential energy at the planet's surface (final position) = - (6.67 x 10^-11) x (5 x 10^29) x (1 x 10^4) / (5 x 10^7)

2. Next, we can find the initial kinetic energy of the meteorite. The kinetic energy is given by:

Kinetic energy at infinity (initial position) = 1/2 mv^2

Where m is the mass of the meteorite (1 x 10^4 kg), and v is the initial speed of the meteorite (3 x 10^2 km/s or 3 x 10^5 m/s).

Kinetic energy at infinity (initial position) = 1/2 x (1 x 10^4) x (3 x 10^5)^2

3. Since mechanical energy is conserved, the total mechanical energy at the initial position (infinity) is equal to the total mechanical energy at the final position (planet's surface). Therefore, we can equate the potential and kinetic energies to find the final speed of the meteorite.

Potential energy at the planet's surface (final position) = Kinetic energy at infinity (initial position)

- (6.67 x 10^-11) x (5 x 10^29) x (1 x 10^4) / (5 x 10^7) = 1/2 x (1 x 10^4) x (3 x 10^5)^2

4. Now, we can solve for the final speed of the meteorite.

Rearranging the equation:

(3 x 10^5)^2 = [2 x (6.67 x 10^-11) x (5 x 10^29) x (1 x 10^4)] / (5 x 10^7)

Taking the square root of both sides:

3 x 10^5 = sqrt{[2 x (6.67 x 10^-11) x (5 x 10^29) x (1 x 10^4)] / (5 x 10^7)}

Calculating the right-hand side:

3 x 10^5 = 1.43 x 10^4 m/s (approximately)

Therefore, the speed of the meteorite just before it hits the surface of the planet is approximately 1.43 x 10^4 m/s.