Gravitational Potential Energy - Kinetic Energy

[Curveball]

Homework Statement

An asteroid of mass m = 2.6 × 10^6 kg approaches the Earth. When it is exactly 3 Earth radii away from the Earth's centre its speed relative to the Earth's centre is u = 8.7 × 10^3 m s–1.

The asteroid subsequently falls down to the Earth's surface. You may assume that the asteroid remains intact and does not dissipate any energy as it passes through the Earth's atmosphere. You may also ignore the rotation of the Earth and of the asteroid.

Calculate the kinetic energy of the asteroid just before it hits the ground. Give your answer by entering a number, specified to an appropriate number of significant figures, in the empty box below.

Homework Equations

Energy Gravitational = -GMm/r
The Earth has mass ME = 5.98 × 10^24 kg and radius 6.38 × 10^6 m.

To find the relevant potential energies you will need to use G = 6.67 × 10^–11 N m^2 kg^–2

The Attempt at a Solution

[/B]
I believe the best method here is to find the Energy Gravitational when the asteroid is 3 radii away form the centre of Earth and when it is just above the Earth's surface.

The difference is the change in potential energy which is then the kinetic energy gained by the asteroid.

E grav change using -GMm/r = (5.41824 x 10^13 ) - (1.625472727 X 10^14) = 1.083648 x 10^14 Joules

Also the kinetic energy at the start would be 0.5mu^2 = 0.5 (2.6 × 10^6 )(8.7 × 10^3)^2 = 9.84 x 10^13 Joules.

Adding the 2 gives kinetic energy equal to 2.0673 X 10^14 Joules.

 

[RUber]

Sounds good to me. I checked the calculations and all look correct.
You are using $(Egrav_1 + Ekin_1)=(Egrav_2 + Ekin_2)$.