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.

   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 ◄

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

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

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