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.

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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft