Question

A satellite is in a circular orbit around a planet. The ratio of the satellite's kinetic energy to its gravitational potential energy, \(K / U_{\mathrm{g}}\), is a constant whose value is independent of the masses of the satellite and planet, and of the radius and velocity of the orbit. Find the value of this constant. (Potential energy is taken to be zero at infinite separation.)

Short answer

Based on the step-by-step solution provided, the constant ratio of a satellite's kinetic energy to its gravitational potential energy is -2. This ratio is independent of the masses of the satellite and planet, as well as the radius and velocity of the orbit.

Step-by-step solution

Write down the formulas for kinetic energy and gravitational potential energy

The formula for kinetic energy is \(K = \frac{1}{2}m_{\text{s}}v^2\), where \(m_{\text{s}}\) is the mass of the satellite and \(v\) is its orbital velocity. The formula for gravitational potential energy is \(U_\text{g} = -\frac{Gm_{\text{p}}m_{\text{s}}}{r}\), where \(m_{\text{p}}\) is the mass of the planet, \(r\) is the orbital radius, and \(G\) is the gravitational constant.

Find a relationship between the orbital velocity and the satellite's mass and radius

We can use the centripetal force formula, which states that the force required to keep an object moving in a circle is given by \(F_\text{c} = \frac{mv^2}{r}\). The gravitational force between the planet and satellite is what provides the centripetal force in this case, so we have \(F_\text{g} = G\frac{m_{\text{p}}m_{\text{s}}}{r^2}\). Setting the two equal, we get the following relationship: $$\frac{m_{\text{s}}v^2}{r} = G\frac{m_{\text{p}}m_{\text{s}}}{r^2}$$

Solve for orbital velocity squared

We now have one equation with one unknown, \(v^2\). We can solve this equation for \(v^2\) by multiplying both sides by \(r^2\) and dividing both sides by \(m_{\text{s}}\): $$v^2 = \frac{Gm_{\text{p}}r}{m_{\text{s}}}$$

Replace orbital velocity squared in the kinetic energy equation

Now that we have a relationship for \(v^2\), we can replace it in the kinetic energy equation to eliminate the dependency on the satellite's mass and radius: $$K = \frac{1}{2}m_{\text{s}}\left(\frac{Gm_{\text{p}}r}{m_{\text{s}}}\right)$$

Simplify the kinetic energy equation

Cancel out the common term of \(m_{\text{s}}\) from the equation: $$K = \frac{1}{2}Gm_{\text{p}}r$$

Divide the kinetic energy by the gravitational potential energy

Now we can divide the kinetic energy equation by the gravitational potential energy equation to find the ratio: $$\frac{K}{U_{\mathrm{g}}} = \frac{\frac{1}{2}Gm_{\text{p}}r}{-\frac{Gm_{\text{p}}m_{\text{s}}}{r}}$$

Simplify the ratio equation

Cancel out common terms and simplify the ratio: $$\frac{K}{U_{\mathrm{g}}} = \frac{2}{-m_{\text{s}}}$$ Since the value of the ratio is independent of the mass of the satellite, we can conclude that the constant ratio is: $$\frac{K}{U_{\mathrm{g}}} = -2$$

Key concepts

Kinetic Energy in Circular Orbits

Kinetic energy represents the energy an object has due to its motion. For a satellite moving in a circular orbit, this is crucial as it maintains the satellite's consistent path around a planet. The formula to calculate the kinetic energy (\(K\)) is \(K = \frac{1}{2}m_{\text{s}}v^2\), where \(m_{\text{s}}\) is the satellite’s mass, and \(v\) is its velocity.

Our step by step solution shows that we can derive the velocity from the centripetal force that acts upon the satellite. This is especially fascinating in space dynamics as it links the concept of motion directly with the force of gravity. Understanding kinetic energy in the context of circular orbits helps us visualize how a satellite is able to maintain its height and speed without either falling into the planet below or drifting off into space.

Gravitational Potential Energy in Space

Gravitational potential energy (\(U_{\mathrm{g}}\)) is the energy an object possesses due to its position within a gravitational field. The formula is \(U_\text{g} = -\frac{Gm_{\text{p}}m_{\text{s}}}{r}\), where \(G\) is the universal gravitational constant, \(m_{\text{p}}\) is the planet's mass, \(m_{\text{s}}\) is the satellite's mass, and \(r\) is the radius of the orbit. The negative sign indicates that the energy is more negative closer to the planet, and energy is considered zero at infinite separation.

Gravitational potential energy plays an essential role in celestial mechanics, as it accounts for the non-linear relationship between distance and force. As a teaching tip, it's important to highlight that potential energy is at a maximum when a satellite is farthest from the attraction source, and minimum when it is closest, inverting our usual perceptions of 'high' and 'low' energy states.

Centripetal Force and Orbital Stability

Centripetal force is what keeps an object moving in a circular path, acting perpendicular to its motion towards the center of rotation. For an object in a circular orbit, such as our satellite, this force is provided by gravity. The formula for centripetal force \(F_\text{c}\) is \(F_\text{c} = \frac{mv^2}{r}\), with \(m\) being the mass of the object, \(v\) its orbital velocity, and \(r\) the radius of the orbit.

In the satellite’s circular orbit, the gravitational force acts as the centripetal force, which links the speed of the satellite to the gravitational pull of the planet. Through this relationship, students can grasp how the satellite's speed must be precisely calculated to keep the orbit stable. A satellite with too much kinetic energy would escape the orbit, while one with insufficient kinetic energy would spiral into the planet. Therefore, the delicate balance of centripetal force is what allows satellites to maintain their orbits for extended periods.