Respuesta :
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
Answer:
K2 = 4/3 K1
Explanation:
The objects are initially at rest and gravity is a conservative force. Therefore, the kinetic energy of each object just before hitting the surface of Earth is equal and opposite to the change in the object's gravitational potential energy.
Let E and E denote mass and radius of Earth, respectively, and let denote mass of the objects. In Case 1, the height of the object above Earth's surface is E. In Case 2, the height of the object above Earth's surface is 2E. Assume that the entire mass of Earth is concentrated at its center, and use the general expression for gravitational potential to write the equations for the change in potential energy Δ for both cases.
Δ1=−E(1E−1E+E)=(−1/2) E/E
Δ2=−E(1E−1E+2E)=(−2/3) E/E
Next, use the fact that the change in kinetic energy equals the change in potential energy. Divide one equality by the other, then solve for 2.
2/1=Δ2/Δ1=(−2/3)(E)/(E) / (−1/2)(E)/E=(2/3) / (1/2)=(4/3)
Therefore, 2=4/31.
