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.
Compare the kinetic energy of the two objects just before they hit the surface of the earth.
K2 = 2 K1
K2 = 4 K1
K2 = (4/3) K1
K2 = (3/2) K1

Answer 1

The kinetic energies just before touching the ground are as follows;

Case 1

`W_(1) = G* M_(E)* m * (1)/(2 * r_(2))`

Case 2

`W_(2) = G* M_(E)* m * (2)/(3 * r_(2))`

The correct option is;

K2 = (4/3) K1

Step-by-step explanation:

The work done is given by the relation;

For case 1

`W_1 = W_(1\rightarrow 2) = G* M_(E)* m * \left ((1)/(r_(2))-(1)/(r_(1)) \right )`

Where for case 1 we have:

G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r₂ = Radius of the Earth

r₁ = 2 × Radius of the Earth = 2 × r₂

Hence;

`W_2 = W_(1\rightarrow 2) = G* M_(E)* m * \left ( (1)/(r_(2)) - (1)/(2* r_(2)) \right )`

`W_(1\rightarrow 2) = G* M_(E)* m * (1)/(2 * r_(2))`

For case 2 we have:

G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r₂ = Radius of the Earth

r₁ = 3 × Radius of the Earth = 2 × r₂

Hence;

`W_(1\rightarrow 2) = G* M_(E)* m * \left ( (1)/(r_(2)) - (1)/(3* r_(2)) \right )`

`W_(1\rightarrow 2) = G* M_(E)* m * (2)/(3 * r_(2))`

Therefore;

`(\Delta K2)/(\Delta K1) = (W_(2))/(W_(1)) = (G* M_(E)* m * (2)/(3 * r_(2)))/(G* M_(E)* m * (1)/(2* r_(2))) = (4)/(3)`

Hence;

K2 = (4/3) × K1.