A satellite is in circular orbit around the Earth at a distance of 4.74RE with a speed of 3.63 km/s. With what speed would the satellite hit the Earth's surface if somehow it suddenly stopped and fell to Earth? Ignore air resistance.
To determine the speed at which the satellite would hit the Earth's surface, we can use the principle of conservation of mechanical energy.
The initial state of the satellite is when it is in circular orbit around the Earth. In this state, the satellite has both kinetic energy (KE) due to its velocity and potential energy (PE) due to its distance from the center of the Earth. The formula for mechanical energy (E) is given by:
E = KE + PE
We can ignore the potential energy at the Earth's surface as it is considered zero. Therefore, the mechanical energy of the satellite in orbit is equal to its kinetic energy.
Next, we need to calculate the satellite's initial kinetic energy. The formula for kinetic energy is given by:
KE = 1/2 * m * v^2
Where m is the mass of the satellite and v is its velocity. Since the mass of the satellite is not given, we can ignore it for now and focus on the ratio of the kinetic energies.
Now, let's consider the final state when the satellite falls to the Earth's surface. In this state, the satellite has lost its gravitational potential energy, but its kinetic energy remains the same. However, the satellite's velocity would have changed since it stopped suddenly. Let's call this final velocity vf.
Since the kinetic energy is conserved, we have:
KE_initial = KE_final
Substituting the kinetic energy equations for both initial and final states:
1/2 * m * v_initial^2 = 1/2 * m * vf^2
Now, let's solve for vf:
vf^2 = v_initial^2
vf = sqrt(v_initial^2)
Where v_initial is the initial velocity of the satellite in orbit. Substituting the given value, vf can be calculated as:
vf = sqrt((3.63 km/s)^2)
vf ≈ 3.63 km/s
Therefore, if the satellite were to suddenly stop and fall to Earth, it would hit the Earth's surface with a speed of approximately 3.63 km/s.