Question

Consider two identical objects released from rest high above the surface of the earth (neglect air resistance for this question). In Case 1 we release an object from a height above the surface of the earth equal to 1 earth radius, and we measure its kinetic energy just before it hits the earth to be K1. In Case 2 we release an object from a height above the surface of the earth equal to 2 earth radii, and we measure its kinetic energy just before it hits the earth to be K2. 1)Compare the kinetic energy of the two objects just before they hit the surface of the earth.

Answer 1

K2 = 4/3 K1

Step-by-step 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.

Answer 2

the relationship between the two scientific energies is K2 / K1 = 8/9

Step-by-step 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