physics question?

It gains energy = mGM/r. You need to replace GM. m is the mass of the satellite For this use
g=GM/R²(earth)
Energy gain will be= m gR²(Earth)/R.
Velocity added will be = sqrt(2gR²(earth)/R)
We know R=4R(erath). You get
v= sqrt(2 g /4 *R(Earth))
R=7000km, g= 9.8/1000 km
v=11.7km/s. This is added to the satellite. The velocity will be 10+11.7= 21.7 km/s....Show more

 
   F = -G.m.M/r² = -m.g
> g = G.M/r² = [G.M/R²]*[1/(1+z/R)]² = g○/(1+z/R)²
   F = -dEp/dr
> Ep = -G.m.M/r = r.F = -(R + z).m.g = -R.(1+z/R).m. g○/(1+z/R)² = -R.m.g○/(1+z/R)

   R = 6 371.0 km : average Earth ray in the spherical approximation
   go = 9.806 65 N/kg : gravity on superficy
   z : altitude = 3.R = h

   Em = Ep + Ec = cste
> -R.m.g○/(1+h/R) + m.V○²/2 = -R.m.g○ + m.V²/2
> V² = V○² + 2.R.g○(1 - 1/(1+h/R))
   1 - 1/(1+h/R) = h/R/(1+h/R)
> V² = V○² + 2.g○.h/(1+h/R)

> V = (1E8 + 2*9.807*3*6.37E6/(1 + 3))^0.5 = 13918 m/s ≈ 14 km/s ◄
 ...Show more

Zero. Once it hits the earth it stops.

Immediately prior to hitting the earth it will have reduced to around 1/10 km/s due to the resistance of the atmosphere.
Although the speed will depend on the size and shape of the rocket.

You cannot treat such a fast moving object as a ballistic mass.

Of course if you eliminate the atmosphere first then you have
g = GM/Re^2
U2 = - GMm/ Re = - gm* Re
U1 = - GMm/ 4Re = - gm * Re/ 4
So the gain in kinetic energy = 3gm * Re/4
ie ek2 = Ek1 + 3gm*Re/4
1/2 m v2^2 = 1/2 m V1^2 + 3gm* Re/4
v2^2 = v1^2 + 6g* Re/4
v2 = sqrt( v1^2 + 6g * Re/4)
= sqrt( 10^8 + 6*9.8*6378*10^3) = 21.8 km/s...Show more

Zero, if it properly executes the reentry and landing burns....Show more

Assuming no atmosphere friction, which is far from the actual facts.

Gravitational potential energy (to center of earth)
E = G m₁m₂/r
Gravitational potential energy (to surface)
from height h
E = [GmM/(R+h)] – [GmM/R]
E = GmM[(1/(R+h)) – (1/R)]
earth radius R = 6,371 km = 6.37e6 meters
G = 6.674e-11 m³/kgs²
earth mass M 5.974e24 kg

"spotted 4 times the radius of the earth away" assuming that is measured from the surface, as that is where the spotters would be.

E = 6.674e-11(m)5.974e24[(1/(6.37e6 +4• 6.37e6)) – (1/6.37e6)]
E = 6.674e-11(m)5.974e18[(1/(5• 6.37)) – (1/6.37)]
E = 6.674e-11(m)5.974e18[0.1255887]
E = m•5.01e7 J

now add the KE from the 10000 m/s, KE = ½mV² = ½m1e4² = ½m1e8 = m•5e7

total is m•10e7 J

when it hits, all that will be velocity
KE = ½mV² = m•10e7
V² = 5e7
V = 7070 m/s...Show more