A meteorite of mass m= 2 ×10^4 kg is approaching head-on a planet of mass M= 6 ×10^29 kg and radius R= 8 ×10^4 km. Assume that the meteorite is initially at a very large distance from the planet where it has a speed v0= 7 ×10^2 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 principle of conservation of energy.
The initial total mechanical energy of the system (meteorite + planet) is equal to the final total mechanical energy just before impact.
The initial total mechanical energy can be calculated as the sum of the kinetic energy and the gravitational potential energy of the meteorite at a very large distance from the planet:
Ei = (1/2)mv0^2 + (-GMm/R)
Here, Ei is the initial total mechanical energy, m is the mass of the meteorite, v0 is the initial speed of the meteorite, G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.
The final total mechanical energy just before impact is equal to the kinetic energy of the meteorite:
Ef = (1/2)mv^2
Here, Ef is the final total mechanical energy just before impact, and v is the speed of the meteorite just before it hits the surface.
Since there is no change in the gravitational potential energy of the meteorite during its motion towards the planet, the initial total mechanical energy is equal to the final total mechanical energy:
Ei = Ef
(1/2)mv0^2 + (-GMm/R) = (1/2)mv^2
Now, let's solve this equation for v.
First, let's convert all the given values into SI units for consistency.
Mass of the meteorite, m = 2 × 10^4 kg
Mass of the planet, M = 6 × 10^29 kg
Radius of the planet, R = 8 × 10^7 m
Initial speed of the meteorite, v0 = 7 × 10^5 m/s
Gravitational constant, G = 6.67 × 10^-11 m^3/kg/s^2
Substituting these values into the equation, we have:
(1/2)(2 × 10^4 kg)(7 × 10^5 m/s)^2 + (-6.67 × 10^-11 m^3/kg/s^2)(2 × 10^4 kg)(6 × 10^29 kg)/(8 × 10^7 m) = (1/2)(2 × 10^4 kg)v^2
Simplifying and solving for v, we get:
7 × 10^11 + -5.0024 × 10^11 = v^2
v^2 = 11.0024 × 10^11
v = √(11.0024 × 10^11)
Calculating this value, we find that v ≈ 3.32 × 10^6 m/s.
Therefore, the speed of the meteorite just before it hits the surface of the planet is approximately 3.32 × 10^6 m/s.