A meteorite of mass m= 1 ×104 kg is approaching head-on a planet of mass M= 3 ×1029 kg and radius R= 7 ×104 km. Assume that the meteorite is initially at a very large distance from the planet where it has a speed v0= 2 ×102 km/s. Take G= 6.67 ×10−11.

Determine the speed of the meteorite v (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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To determine the speed of the meteorite just before it hits the surface of the planet, we can use the law of conservation of energy.

The initial potential energy of the meteorite when it is at a very large distance from the planet can be given as:

PE_initial = - (G * M * m) / R

where G is the gravitational constant, M is the mass of the planet, m is the mass of the meteorite, and R is the radius of the planet.

The initial kinetic energy of the meteorite at this point is given by:

KE_initial = (1/2) * m * v0^2

where v0 is the initial velocity of the meteorite.

At the surface of the planet, just before impact, the potential energy of the meteorite is zero because it is at the surface. Therefore, all the initial potential energy is converted into kinetic energy.

So, the final kinetic energy of the meteorite just before impact is:

KE_final = PE_initial

Setting the two expressions for kinetic energy equal, we have:

(1/2) * m * v0^2 = -(G * M * m) / R

Now, we can solve for the final velocity (v) by rearranging the equation:

v^2 = -2 * (G * M) / R + v0^2

Taking the square root of both sides, we get:

v = sqrt(-2 * (G * M) / R + v0^2)

Now, we can plug in the given values and calculate the final velocity:

v = sqrt(-2 * (6.67 ×10^-11 * 3 ×10^29) / (7 ×10^4) + (2 ×10^2)^2)

v ≈ 1.98 ×10^4 m/s

Therefore, the speed of the meteorite just before it hits the surface of the planet is approximately 1.98 ×10^4 m/s.