A satellite with a mass of 284 kg approaches a large planet at a speed vi,1 = 14.4 km/s. The planet is moving at a speed vi,2 = 11.0 km/s in the opposite direction. The satellite partially orbits the planet and then moves away from the planet in a direction opposite to its original direction (see the figure). If this interaction is assumed to approximate an elastic collision in one dimension, what is the speed of the satellite after the collision?
To determine the speed of the satellite after the collision, we need to take into account the conservation of momentum and the conservation of kinetic energy. Let's break down the problem into steps:
Step 1: Convert all given speeds to the same unit
The speed of the satellite vi,1 is given as 14.4 km/s, and the speed of the planet vi,2 is given as 11.0 km/s. Let's convert both speeds to m/s to keep the units consistent.
vi,1 = 14.4 km/s * (1000 m/km) / (3600 s/h)
= 4 m/s
vi,2 = 11.0 km/s * (1000 m/km) / (3600 s/h)
= 3.06 m/s
Step 2: Apply the conservation of momentum
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Momentum before collision = Momentum after collision
(mass of satellite * initial velocity of satellite) + (mass of planet * initial velocity of planet) = (mass of satellite * final velocity of satellite) + (mass of planet * final velocity of planet)
(284 kg * 4 m/s) + (unknown * 3.06 m/s) = (284 kg * unknown) + (unknown * -3.06 m/s)
Step 3: Solve the equation for the final velocity of the satellite
Simplifying and rearranging the equation, we can solve for the unknown final velocity of the satellite.
(284 kg * 4 m/s) + (3.06 m/s * unknown) = (284 kg * unknown) - (3.06 m/s * unknown)
1136 kg·m/s + 3.06 m/s * unknown = 284 kg * unknown - 3.06 m/s * unknown
1136 kg·m/s = (284 kg - 3.06 kg) * unknown
1136 kg·m/s = (280.94 kg) * unknown
unknown = 1136 kg·m/s / 280.94 kg
unknown ≈ 4.05 m/s
Therefore, the speed of the satellite after the collision is approximately 4.05 m/s.