At a location 4Re away from the center of the Earth, you release a 2kg object from rest. How fast is the object traveling when it is 2Re away from the center of the Earth?

Re = 6.4 x 10^6 m
Me = 6 * 10^24 m

PEi = - GMem/4Re

= KEf + PEf
= KEf + [- GMem/2Re]
So KEf = GMem/2Re - GMem/4Re
or (1/2)m v^2 = GMem/2Re
v^2 = G*Me/Re

To find the speed of the object, we can use the principle of conservation of mechanical energy. At the initial position, where the object is 4Re away from the Earth's center, it has potential energy (PE) only. As the object falls towards the center, potential energy is converted into kinetic energy (KE). At the final position, where the object is 2Re away from the center, all the initial potential energy is converted into kinetic energy.

The potential energy of an object at a distance r from the center of the Earth is given by the equation:

PE = G * Me * m / r,

where G is the gravitational constant, Me is the mass of the Earth, m is the mass of the object, and r is the distance from the center of the Earth.

Since the object is released from rest, its initial kinetic energy (KEi) is zero. At the final position, its potential energy is completely converted into kinetic energy, so:

PE = KEf.

Using the equation for potential energy, we can write:

G * Me * m / (4Re) = 0.5 * m * v^2,

where v is the velocity of the object at a distance of 2Re from the center.

Simplifying the equation by canceling out the mass (m) and rearranging, we get:

v^2 = 2 * G * Me / (4Re),

v^2 = G * Me / (2Re),

v = √(G * Me / (2Re)).

Now we can plug in the values:

G = 6.67 x 10^-11 N m^2/kg^2,

Me = 6 x 10^24 kg,

Re = 6.4 x 10^6 m.

Calculating the expression:
v = √(6.67 x 10^-11 N m^2/kg^2 * 6 x 10^24 kg / (2 * 6.4 x 10^6 m))

v ≈ 11127 m/s.

Therefore, when the object is 2Re away from the center of the Earth, it is traveling at approximately 11127 m/s.