Question

For the satellite in Solved Problem 12.2, orbiting the Earth at a distance of \(3.75 R_{\mathrm{E}}\) with a speed of \(4.08 \mathrm{~km} / \mathrm{s}\) with what speed would the satellite hit the Earth's surface if somehow it suddenly stopped and fell to Earth? Ignore air resistance.

Short answer

Answer: Approximately 3352 m/s.

Step-by-step solution

Calculate the initial potential and kinetic energy of the satellite in its orbit

The satellite in orbit has potential energy due to Earth's gravity, given by \(U = -\frac{Gm_{s}m_{E}}{r}\), where \(G\) is the gravitational constant, \(m_{s}\) is the mass of the satellite, \(m_{E}\) is the mass of the Earth, and \(r\) is the distance between the center of the Earth and the center of the satellite. The kinetic energy of the satellite is given by \(K = \frac{1}{2}m_{s}v^2\), where \(v\) is the speed of the satellite in its orbit (\(4.08 \mathrm{~km/s} = 4080 \mathrm{~m/s}\)). Since we don't know the exact mass of the satellite, we will temporarily ignore \(m_{s}\) and include it back in later calculations.

Calculate the final potential energy and relate it to the change in kinetic energy

When the satellite reaches the Earth's surface, its potential energy will be \(U_{\text{final}} = -\frac{Gm_{s}m_{E}}{R_{\text{E}}}\), where \(R_{\text{E}}\) is the Earth's radius. The change in potential energy is equal to the change in kinetic energy, which can be written as \(\Delta U = \Delta K\). Therefore, we have: \(-\frac{Gm_{s}m_{E}}{R_{\text{E}}} + \frac{Gm_{s}m_{E}}{3.75R_{\text{E}}} = \frac{1}{2}m_{s}v_\text{final}^2 - \frac{1}{2}m_{s}v_\text{initial}^2\)

Solve for the final speed of the satellite

Now we can solve for \(v_\text{final}\): \(v_\text{final}^2 = \frac{2Gm_{E}}{R_{\text{E}}} - \frac{2Gm_{E}}{3.75R_{\text{E}}} + v_\text{initial}^2\) \(v_\text{final}^2 = \frac{2Gm_{E}}{3.75R_{\text{E}}} + 4080^2\) We can plug in the values for the gravitational constant \(G\) (\(6.674 \times 10^{-11} \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}\)) and Earth's mass \(m_{E}\) (\(5.972 \times 10^{24} \mathrm{kg}\)), and radius \(R_{E}\) (\(6.371 \times 10^6 \mathrm{m}\)) \(v_\text{final}^2 = \frac{2 \times (6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{3.75(6.371 \times 10^6)} + 4080^2\) \(v_\text{final}^2 = 1.122\times 10^{7}\) Taking the square root, we get: \(v_\text{final} = \sqrt{1.122\times 10^{7}}\) \(v_\text{final} \approx 3352 \mathrm{~m/s}\) So, if the satellite suddenly stopped and fell to Earth, it would hit the Earth's surface with a speed of approximately \(3352 \mathrm{~m/s}\).

Key concepts

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. Think of it as the energy stored when you lift an object against gravity. For a satellite in orbit around Earth, GPE is negative, which may seem odd at first. This is because the reference point is taken at an infinite distance away, where the gravitational potential energy is considered to be zero. The formula for GPE is given by \( U = -\frac{Gm_{s}m_{E}}{r} \), where \(G\) is the gravitational constant, \(m_{s}\) is the satellite's mass, \(m_{E}\) is Earth's mass, and \(r\) is the distance from the Earth's center to the satellite. The negative sign shows that energy is released as the satellite falls toward Earth.

To get a better grasp, imagine you're holding a rock above the ground. The higher you hold it, the greater the GPE. If you let go, the rock falls as the GPE is converted to kinetic energy. In the case of our satellite, if it's suddenly not in orbit but stationary relative to Earth, it would fall back to Earth and all its GPE would convert into kinetic energy by the time it hits the ground.

Kinetic Energy in Orbit

Kinetic energy (KE) is the energy of motion. Anything that moves has kinetic energy, which is calculated using the formula \( K = \frac{1}{2}m_{s}v^2 \), where \( m_{s} \) is the mass of the moving object and \( v \) is its velocity. For satellites, the kinetic energy keeps them in orbit; it is the speed they need to balance the pull of gravity so they neither drift away into space nor fall back to Earth.

Imagine spinning a ball on a string in a circular motion. The ball tries to move in a straight line due to its kinetic energy, but the string's tension (akin to gravity for the satellite) keeps it moving in a circle. The satellite in our exercise has a lot of kinetic energy while in orbit at 4.08 km/s. If that velocity suddenly went to zero, gravity would take over completely, pulling the satellite back to Earth.

Conservation of Mechanical Energy

The law of conservation of mechanical energy states that the total mechanical energy (the sum of potential and kinetic energy) in a system remains constant if only conservative forces (like gravity) are doing work. In the context of a satellite orbiting Earth, even though the satellite's potential and kinetic energy are continually being converted into each other, their sum remains the same.

In our satellite's plunge towards Earth, we assume no air resistance to simplify the calculation. Therefore, the loss of gravitational potential energy as the satellite fall must equal the gain in kinetic energy. By setting the initial potential and kinetic energy equal to the final kinetic energy (since the final potential energy at Earth's surface is a constant value), we establish the conservation of mechanical energy. Using this principle, we calculated the speed at which the satellite would hit the Earth, allowing us to understand the satellite's energy transformation throughout its extreme fall back to Earth.