What is the speed of a space probe when it is very far from earth if it was launched from the surface of earth at twice its escape speed?

V = 20,928.32 m/s

Explanation:

First, we need to know the escape speed. As the space probe it's launched from the surface of Earth, the escape speed from Earth is about 11,186 m/s, so twice is 22,372 m/s.

We now have the innitial speed.

To solve this, we will use the expression of conservation of energy which is:

K1 + U1 = K2 + U2

Where:

K: kinetic energy (1/2 mV²)

U: potential energy (-GmM/r)

We will say that 1 it's the energy from the earth, and 2, when it's far away from earth.

Now, since we want the speed of the probe at very far distances from the Earth, we set the distance to be infinity, so equation (1) becomes:

1/2 mV1² - GmM/r1 = 1/2 mV2² - GmM/r∞ (2)

And since the gravitational potential energy is inversely proportional to r, then the gravitational potential energy will thus become zero, therefore, equation (2) will become in:

1/2 mV1² - GmM/r1 = 1/2 mV2² (3)

From here, we can cancel both m, and solve for V2:

1/2 V1² - GM/r1 = 1/2 V2²

V2² = V1² - 2GM/r1

V2 = √V1² - 2GM/r1 (4)

Where:

G: Gravitational force (6.67x10^-11 m³/kg²)

M: mass of Earth (5.97x10^24 kg)

r: Radius of Earth (6,37x10^6 m)

Replacing this values and the escape speed, we can solve for the speed of the probe very far away from Earth

V2 = √ (22,372) ² - (6.67x10^-11 * 5.97x10^24 / 6.37x10^6)

V2 = √437,994,767

V2 = 20,928.32 m/s