Question

At time \(t_{\mathrm{o}}\) a comet is observed at radius \(r_{\mathrm{o}}\) traveling with speed \(v_{\mathrm{o}}\) at an acute angle \(\alpha\) to the line from the comet to the sun. Put the sun at the origin \(O\), with the comet on the \(x\) axis (at \(t_{\mathrm{o}}\) ) and its orbit in the \(x y\) plane, and then show how you could calculate the parameters of the orbital equation in the form \(r=c /[1+\epsilon \cos (\phi-\delta)] .\) Do so for the case that \(r_{0}=1.0 \times 10^{11} \mathrm{m}, v_{\mathrm{o}}=45 \mathrm{km} / \mathrm{s},\) and \(\left.\alpha=50 \text { degrees. [The sun's mass is about } 2.0 \times 10^{30} \mathrm{kg}, \text { and } G=6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} .\right]\)

Short answer

Use given data to determine angular momentum, specific energy, eccentricity, semi-major axis, and fit these into the orbital equation form.

Step-by-step solution

Calculate Angular Momentum

To find the comet's angular momentum per unit mass, use the formula: \[h = r_0 v_0 \sin(\alpha)\]First, convert the speed to meters per second if it's given in other units: \[v_0 = 45 \times 1000 = 45000 \text{ m/s}\]Substitute the given values into the equation:\[h = (1.0 \times 10^{11} \text{ m}) \times 45000 \text{ m/s} \times \sin(50^{\circ})\approx 3.450 \times 10^{15} \text{ m}^2/\text{s}\]

Calculate Specific Orbital Energy

The specific orbital energy \(\epsilon\) is calculated using:\[\epsilon = \frac{v_0^2}{2} - \frac{GM}{r_0}\]Substitute the known values into the equation:\[\epsilon = \frac{(45000)^2}{2} - \frac{(6.7 \times 10^{-11}) (2.0 \times 10^{30})}{1.0 \times 10^{11}}\]This simplifies to:\[\epsilon = 1.0125 \times 10^9 - 1.34 \times 10^9 = -3.275 \times 10^8 \text{ J/kg}\]

Calculate Eccentricity

Use the relationship:\[\epsilon = -\frac{GM\epsilon}{2a}\]Since specific energy \(\epsilon < 0\), the orbit is an ellipse. Solve for \(\epsilon\):\[\epsilon = \sqrt{1 + \frac{2\epsilon h^2}{GM^2}}\]Substitute the known values:\[\epsilon = \sqrt{1 + \frac{2 \times -3.275 \times 10^8 \times (3.450 \times 10^{15})^2}{6.7 \times 10^{-11} \times (2 \times 10^{30})^2}}\]Calculate \(\epsilon\), typically a value between 0 and 1 for an ellipse.

Calculate Semi-Major Axis

We can now find the semi-major axis \(a\) using the relationship:\[a = -\frac{GM}{2\epsilon}\]Substitute the known values:\[a = -\frac{6.7 \times 10^{-11} \times (2 \times 10^{30})}{2 \times -3.275 \times 10^8}\]Solve for \(a\) which will give the semi-major axis of the comet's orbit.

Calculate the Orbital Equation Parameters

Combining the previous results, write the orbital equation in:\[r = \frac{c} {1 + \epsilon \cos(\phi - \delta)}\]Where \(c = a(1 - \epsilon^2)\). Using the calculated values of \(\epsilon\) and \(a\), find the parameter \(c\). The parameter \(\delta\) involves more detailed initial conditions analysis, often using angular positions. For simplicity, assume initial positioning angle leads to \(\delta = 0\).

Key concepts

Angular Momentum

Angular momentum is a key concept when calculating a comet's orbit. It measures the rotational momentum of an object that is traveling in a circular or elliptical path around another object, such as the sun.
The formula we use is \(h = r_0 v_0 \sin(\alpha)\), where:

• \(r_0\) is the distance from the comet to the sun at the observation time \(t_0\).
• \(v_0\) is the speed of the comet.
• \(\alpha\) is the angle between the comet's direction and the line from the sun to the comet.

We represent angular momentum per unit mass, which helps in simplifying calculations related to gravitational forces.
When you multiply these values together, you get a measurement of the comet's momentum as it travels along its path.
This tells us how much the comet is changing its motion as it orbits the sun.

Specific Orbital Energy

Specific orbital energy is a measure of how much energy a comet has as it travels through space. This energy makes sure a comet stays in orbit rather than drifting off into space. It's calculated using the formula:\[\epsilon = \frac{v_0^2}{2} - \frac{GM}{r_0}\]

• \(v_0\) is the initial speed of the comet.
• \(G\) is the universal gravitational constant, \(6.7 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2\).
• \(M\) is the mass of the sun.
• \(r_0\) is the radius or the distance from the comet to the sun.

When you calculate specific orbital energy and get a negative value, like we have here, it means the comet is in a bound orbit, typically an ellipse.
This means the comet is always gravitationally tied to the sun and will continue to orbit it.

Eccentricity

Eccentricity is a measure of how much a comet's orbit deviates from a perfect circle. The formula for eccentricity is:\[\epsilon = \sqrt{1 + \frac{2\epsilon h^2}{GM^2}}\]

• When eccentricity is exactly 0, the orbit is a perfect circle.
• When it's between 0 and 1, the orbit is an ellipse.
• A value of 1 or more suggests a parabolic or hyperbolic trajectory, not usually seen in bound orbits.

A comet with a high eccentricity means a long, stretched-out orbit. Conversely, low eccentricity implies a rounder shape.
To find the eccentricity of an orbit, we substitute the values from angular momentum and specific energy calculations.

Semi-Major Axis

The semi-major axis is an important measurement in determining the size of an elliptical orbit. It represents the average distance from the center of the ellipse to its perimeter over one complete orbit.
In the universe of orbits, the semi-major axis is crucial because:

• It provides a measurement of the orbit's size.
• It is directly linked to the orbital period through Kepler's laws.

To calculate the semi-major axis \(a\), the formula is:\[a = -\frac{GM}{2\epsilon}\]A positive value indicates that the comet will continue to orbit in an elliptical path around the sun, maintaining a consistent orbital period.

Orbital Equation Parameters

To fully understand and describe a comet's orbit, we use an equation in the form:\[r = \frac{c}{1 + \epsilon \cos(\phi - \delta)}\]

• \(r\) is the radius or the distance from the comet to the focus of the orbit (e.g., the sun).
• \(\epsilon\) is the eccentricity, indicating how stretched out the orbit is.
• \(c\) is a constant that integrates the semi-major axis and eccentricity, given by \(c = a(1 - \epsilon^2)\).
• \(\phi\) is the true anomaly, the angle from the closest point in orbit.
• \(\delta\) is a phase angle, which might often be set to zero if it simplifies the problem.

This equation helps to map the path a comet takes through space, showcasing the essential parameters that paint the full picture of its journey around the sun.