Answer:
the relationship between the two scientific energies is K2 / K1 = 8/9
Explanation:
They ask us to compare the kinetic energies, so we must use the energy conservation theorem, let's start calculating the gravitational potential energy, to use the universal gravitation equation
F = G m1 m2 / R²
With the mass m1 the Earth mass and m2 the mass of the object, R the distance from the center of the Terra to the object
Let's calculate the potential energy from the equation
F = - dU / dr
dU = - F dr
∫ dU = - ∫ F dr
Uf - Uo = - (Gme m2) I dr / r²
Uf - Uo = - (Gme m2) (1 /rf - 1 /ro)
Let's see the distances in each case
Case 1. tell us that it is launched from 1 terrestrial radio
R = Re + Re = 2 Re
Case 2. It is released from 2 terrestrial radios
R = Re + 2 Re = 3 Re
Let's calculate the potential energy for each case
Case 1
ΔU = (Gme m2) [1 / (Re + Re) - 1 / Re)] = (Gme m2) 1 / Re [1/2 +1)
ΔU = (G m2 / Re) 3/2 m2
Case 2
ΔU (Gme m2) [1 / (Re + 2Re) + 1 / Re] = (Gme m2) 1 / Re [1/3 + 1]
Δu = (Gme) 1 / Re 4/3 m2
Having the potential energies We can use the energy conservation theorem applied to the initial and final points of the movement.
Em1 = Uo
Em2 = Uf + K
how do they tell us that there is no friction force
Em1 = Em2
Uo = Uf + K
K = Uf -Uo = ΔU
K = ΔU
Let's calculate the kinetic energy for each case
Case 1 r = Re
K1 = (G m2 / Re) 3/2 m2
Case 2 r = 2Re
K2 = (Gme) 1 / Re 4/3 m2
To compare the two energies let's divide one another
K2 / K1 = [(Gme) 1 / Re 4/3 m2] / [= (G m2 / Re) 3/2 m2]
K2 / K1 = (4/3) / (3/2)
K2 / K1 = 8/9
