Question

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?

Answer 1

A space probe launched at twice the escape velocity from Earth will have a final speed equal to the escape velocity when it is very far from the Earth, assuming no other forces act upon it after launch.

Step-by-step explanation:

If a space probe is launched from the surface of Earth at twice its escape velocity, it will still have a significant speed when it is very far from Earth. The escape velocity from Earth is about 11.2 km/s. If a probe is launched at twice this speed, it ignores air resistance and other possible influences like Earth's rotation or orbital motion. To find the final speed when the probe is at a very large distance from Earth, we would need to consider the work done by Earth's gravity in slowing the probe down from twice the escape speed to its final speed.

As the probe moves away from Earth, the pull of Earth's gravity will do negative work on the probe, reducing its kinetic energy. However, when the probe is very far from Earth, the gravitational pull becomes negligible, and thus, the probe will essentially move at constant speed. Without additional forces acting on the probe, by the conservation of energy, the speed at a very large distance will be the same as the excess speed it had beyond the escape velocity. So, if the probe starts at twice the escape velocity, its final speed when far from Earth would be its launch speed minus the escape velocity (2×11.2 km/s - 11.2 km/s = 11.2 km/s).

Answer 2

V = 20,928.32 m/s

Step-by-step 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