A meteorite of mass m= 2 ×104 kg is approaching head-on a planet of mass M= 7 ×1029 kg and radius R= 5 ×104 km. Assume that the meteorite is initially at a very large distance from the planet where it has a speed v0= 5 ×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=

Intiial PE=GMm/r

Initial KE=1/2 m v^2
can you convert that intial energy to final ke?

my final formula is: Vf=sqrt(Vi^2+(mMG/r))

r being the radius of the planet

1/2 m vf^2=1/2 m vi^2+GMm/r

where r is the initial distance of the object. You can ignore the final PE, as the object is very far from the planet initiually. That does not lead to what you got.

but isn't the object initially at r=infinity therefore P_i=0 and the potential energy just before impact Mmg/the radius of the planet? or am i completely of target

1/2 m vf^2=1/2 m vi^2+GMm/r - it's correct!

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.

The total mechanical energy of the system (meteorite + planet) is conserved throughout the motion. At the initial point where the meteorite is at a very large distance from the planet, it has a certain mechanical energy given by:

E_initial = (1/2)mv₀² - (GmM)/(R₀ + ∞),

where m is the mass of the meteorite, v₀ is its initial velocity, M is the mass of the planet, R is the radius of the planet, R₀ is the distance between the meteorite and the planet at the initial point, and G is the gravitational constant.

As the meteorite approaches the planet, its potential energy decreases while its kinetic energy increases. Just before it hits the surface of the planet, the potential energy becomes zero, and the mechanical energy is given by:

E_final = (1/2)mv² - (GmM)/R.

Since the mechanical energy is conserved, E_initial = E_final:

(1/2)mv₀² - (GmM)/(R₀ + ∞) = (1/2)mv² - (GmM)/R.

Since we are neglecting friction and the planet has no atmosphere, we can assume that the meteorite's kinetic energy just before impact is entirely converted into potential energy, making its velocity v just before impact equal to zero. Hence, the equation becomes:

(1/2)mv₀² - (GmM)/(R₀ + ∞) = 0.

To solve for the velocity v, rearrange the equation:

(1/2)mv₀² = (GmM)/(R₀ + ∞).

Now, plug in the given values into the equation:

(1/2)(2 × 10^4 kg)(5 × 10^2 km/s)² = (6.67 × 10^-11 Nm²/kg²)(2 × 10^4 kg)(7 × 10^29 kg)/[(5 × 10^4 km + ∞)].

Simplifying the equation, we get:

(1/2)(2 × 10^4 kg)(5 × 10^5 m/s)² = (6.67 × 10^-11 Nm²/kg²)(2 × 10^4 kg)(7 × 10^29 kg)/(5 × 10^7 m).

Solving this equation will give the speed v of the meteorite just before it hits the surface of the planet in meters per second (m/s).