Question

You have been sent in a small spacecraft to rendezvous with a space station that is in a circular orbit of radius \(2.5000 \cdot 10^{4} \mathrm{~km}\) from the Earth's center. Due to a mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! You do not apply forward thrust in an attempt to chase the station; that would be fatal folly. Instead, you apply a brief braking force against the direction of your motion, to put you into an elliptical orbit, whose highest point is your present position, and whose period is half that of your present orbit. Thus, you will return to your present position when the space station has come halfway around the circle to meet you. Is the minimum radius from the Earth's center-the low point-of your new elliptical orbit greater than the radius of the Earth \((6370 \mathrm{~km}),\) or have you botched your last physics problem?

Short answer

Answer: We must examine the minimum radius of the elliptical orbit, \(r_\text{min}\). If the value of \(r_\text{min}\) is greater than Earth's radius (6370 km), the spacecraft will not crash into Earth. If \(r_\text{min}\) is less than or equal to Earth's radius, it will either crash or be too close to the Earth. The minimum radius can be calculated using the steps outlined in the solution above.

Step-by-step solution

Calculate the period of the circular orbit

To compute the period of the circular orbit, we can use Kepler's third law, which states: \(T^2 = \dfrac{4\pi^2}{GM} a^3\) Where \(T\) is the period, \(G = 6.674 \times 10^{-11} \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}\) is the gravitational constant, \(M = 5.972 \times 10^{24} \mathrm{kg}\) is the mass of the Earth, and \(a\) is the semi-major axis of the circular orbit. Since the initial orbit is circular, the radius is equal to the semi-major axis, so \(a = 2.5000 \cdot 10^{4} \mathrm{~km} = 2.5000 \cdot 10^{7} \mathrm{~m}\). Now we can compute the period \(T\): \(T^2 = \dfrac{4\pi^2}{GM} a^3\) Plugging the values, we get: \(T^2 = \dfrac{4\pi^2}{(6.674 \times 10^{-11} \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}})(5.972 \times 10^{24} \mathrm{kg})} (2.5000 \cdot 10^{7} \mathrm{~m})^3\) Solving for \(T\), we get the period of the circular orbit.

Calculate the semi-major axis of the elliptical orbit

Now we need to find the semi-major axis of the elliptical orbit with half the period of the circular orbit. We can use Kepler's third law again: \(\left(\dfrac{T}{2}\right)^2 = \dfrac{4\pi^2}{GM} a_e^3\) Where \(a_e\) is the semi-major axis of the elliptical orbit. We can plug the previously found period and solve for \(a_e\): \(a_e^3 = \dfrac{GM}{4\pi^2} \left(\dfrac{T}{2}\right)^2\) Using the calculated period value for \(T\), we can find the value for \(a_e\).

Calculate the minimum radius of the elliptical orbit

Once we have the semi-major axis of the elliptical orbit, we can find the minimum radius (closest point to Earth's center) using the relation between the semi-major and semi-minor axes and the ellipse's foci: \(a_e = \dfrac{r_\text{max} + r_\text{min}}{2}\) Where \(r_\text{max}\) is the maximum radius (equal to the initial circular orbit's radius) and \(r_\text{min}\) is the minimum radius we are trying to find. We can solve for \(r_\text{min}\): \(r_\text{min} = 2a_e - r_\text{max}\) Plugging in the values we have found for the semi-major axis, \(a_e\), and the initial circular orbit radius, \(r_\text{max}\), we can find the minimum radius of the elliptical orbit.

Compare the minimum radius to the Earth's radius

Finally, we can compare the minimum radius of the elliptical orbit to the Earth's radius: If \(r_\text{min} > 6370 \mathrm{~km}\), then the minimum radius is greater than Earth's radius, and the spacecraft will not crash into Earth. If \(r_\text{min} \leq 6370 \mathrm{~km}\), then the minimum radius is less than or equal to Earth's radius, and the spacecraft will either crash or be too close to the Earth, botching the last physics problem. Based on the calculated values of \(a_e\) and \(r_\text{min}\), we can determine the outcome.

Key concepts

Kepler's Third Law

Kepler's Third Law of Planetary Motion is pivotal in celestial mechanics and space travel. It states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. For those delving into the specifics of orbital dynamics, this law can be mathematically expressed as

\( T^2 = \dfrac{4\pi^2}{GM} a^3 \)

where \(T\) is the orbital period, \(G\) is the universal gravitational constant, \(M\) is the mass of the central body around which the orbit occurs, and \(a\) is the semi-major axis of the orbit.
For circular orbits, this semi-major axis is identical to the orbit's radius, simplifying calculations.
This law is crucial when planning space missions as it allows commanders to calculate the necessary orbit to rendezvous with other spacecraft. In the case of needing to correct your orbit, understanding Kepler's Law is the first step towards a successful orbital insertion or adjustment.

Circular Orbit

A circular orbit is described as the motion of a body traversing a path with a constant radius around a central point or body. This is a specific case in orbital mechanics where the orbit has zero eccentricity, indicating it is a perfect circle.

In the context of Earth-bound orbits, such as the problem at hand, a spacecraft in a circular orbit maintains a constant speed and altitude above the Earth's surface. The absence of major thruster burns or maneuvers enables a suitably tranquil and predictable flight path. When calculating circular orbits, the semi-major axis \(a\) simplifies to the orbit's radius, easing various calculations and orbital predictions.

Moreover, for orbital rendezvous—the process of meeting another spacecraft in orbit—maintaining a circular path simplifies timing and position strategies as opposed to more complex elliptical orbits where the orbital velocity varies.

Elliptical Orbit

An elliptical orbit, unlike its circular counterpart, varies in distance from the central body at different points along its path. Characterized by its eccentricity, ranging between 0 (which would be a circle) and 1, elliptical orbits feature both a closest point, called the periapsis or perigee when referring to Earth, and a farthest point, known as the apoapsis or apogee for Earth orbits.

The strategic maneuver of shifting a spacecraft from a circular to an elliptical orbit is a common one, especially in scenarios requiring orbital rendezvous or phasing, where timing is critical to meet another object in space. In the textbook example, a spacecraft deliberately enters an elliptical orbit with a specific semi-major axis length to adjust the timing of its orbit relative to a space station.

Achieving an elliptical orbit involves linked variables and parameters such as the semi-major axis, semi-minor axis, and the central body's mass. Grasping the basics of this motion is vital for precision in orbital mechanics and avoiding catastrophic results, like crashing into the Earth when calculated imprecisely.