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

In summary, a spacecraft with an initial circular orbit of $2.5\times {10}^4$ km around Earth is instructed to move to an elliptical orbit with a period half that of the original circular orbit. When calculating the minimum radius of the elliptical orbit, it is found to be approximately 3050 km. This is less than Earth's radius (6370 km), so the spacecraft would crash into Earth.

Step-by-step solution

Calculate the current period of the circular orbit

To calculate the current period of the circular orbit, we will use Kepler's Third Law, which states that $\frac{T^2}{a^3}=\frac{4{\pi }^2}{GM}$, where $T$ is the period of the orbit, $a$ is the distance (semi-major axis) of the orbit, $M$ is the mass of the central body (in this case, Earth), and $G$ is the gravitational constant. Since we are given $a$, we can solve for $T$. The given distance, $a=2.5\times {10}^4$ km, needs to be converted to meters: $a=2.5\times {10}^7$ m. Using the mass of Earth, $M=5.98\times {10}^{24}$ kg, and the gravitational constant, $G=6.674\times {10}^{-11}\frac{{\mathrm{m}}^3}{\mathrm{kg}\cdot {\mathrm{s}}^2}$, we get: $$T^2=\frac{4{\pi }^2}{GM}\times a^3$$ $$T=\sqrt{\frac{4{\pi }^2}{GM}\times (2.5\times {10}^7{)}^3}\approx 5790\text{seconds}$$

Calculate the period of the new elliptical orbit

The problem states that the period of the new elliptical orbit is half that of the original circular orbit. Therefore, we have $T_{\text{ellipse}}=\frac{1}{2}T\approx 2895$ seconds.

Calculate the semi-major axis of the elliptical orbit

Now, we can use Kepler's Third Law again, but this time to find the semi-major axis $a_{\text{ellipse}}$ of the elliptical orbit by solving the equation for $a_{\text{ellipse}}$: $$a_{\text{ellipse}}^3=\frac{GM}{4{\pi }^2}\times T_{\text{ellipse}}^2$$ $$a_{\text{ellipse}}={(\frac{GM}{4{\pi }^2}\times (2895{)}^2)}^{\frac{1}{3}}\approx 2.195\times {10}^7\text{m}$$

Calculate the semi-minor axis of the elliptical orbit

Since in this scenario, the highest point of the ellipse is in the same position as the circular orbit, we know that $a_{\text{ellipse}}-r_{\text{min}}=2.5\times {10}^7\text{m}$ or $r_{\text{min}}=a_{\text{ellipse}}-2.5\times {10}^7\text{m}$. Therefore, $$r_{\text{min}}\approx 2.195\times {10}^7\text{m}-2.5\times {10}^7\text{m}\approx -3.05\times {10}^6\text{m}$$

Calculate the minimum radius of the elliptical orbit

Since the elliptical orbit goes from the maximum to minimum distance around Earth, the minimum radius from the Earth's center, $R_{\text{min}}$, is the absolute value of $r_{\text{min}}$: $$R_{\text{min}}=|-3.05\times {10}^6\text{m}|=3.05\times {10}^6\text{m}\approx 3050\text{km}$$

Compare the minimum radius with Earth's radius

Now we can compare the minimum radius of the new elliptical orbit, $R_{\text{min}}\approx 3050\text{km}$, with Earth's radius $6370\text{km}$. Since $R_{\text{min}}<6370\text{km}$, the spacecraft would not be able to complete the elliptical orbit and would crash into Earth. So, unfortunately, you have botched your last physics problem.

Key concepts

circular orbit

A circular orbit is a path taken by an object revolving around a central body where its distance from the center remains constant. This type of orbit implies that the object maintains a steady speed without needing additional force to change velocity.
In a circular orbit, the gravitational force pulling the satellite towards the center is perfectly balanced by the object's tangential velocity. This delicate balance keeps the satellite in a consistent loop at a constant altitude from the central body.
For any object in a circular orbit, the radius is the key parameter. In our exercise, the circular orbit's radius is given as 25,000 km from Earth's center. Calculating the period of a circular orbit involves using Kepler's Third Law, allowing us to find how long it takes to complete one full orbit. This is crucial when considering maneuvers like moving to an elliptical orbit.

elliptical orbit

An elliptical orbit, unlike a circular orbit, has two focal points, one of which is occupied by the central body. This results in a path where the distance from the orbiting object to the center varies, creating both high and low points called apogee and perigee, respectively.
In this exercise, creating an elliptical orbit was necessary to strategically meet the space station after altering the spacecraft's velocity. The elliptical orbit has a period half that of the original circular orbit, designed to bring the spacecraft and station together.
The shift from a circular to an elliptical orbit is achieved by adding or subtracting velocity at specific points, affecting the orbit's shape. The maximum distance (apogee) coincides with the spacecraft's initial position, while the closest approach (perigee) needed verification to avoid crashing into Earth.

gravitational constant

The gravitational constant, denoted by the symbol "$G$," is a fundamental constant critical in calculating gravitational forces between two masses. It determines the strength of the gravitational pull that objects exert on each other.
In the context of celestial orbits, the gravitational constant is vital for Kepler's Third Law, allowing us to calculate orbital periods and the characteristics of different orbit types. This law relates the square of the orbital period to the cube of the semi-major axis of the orbit.
The value of $G$ is approximately $6.674\times {10}^{-11}{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$. Its precise measurement is essential for accurate orbit predictions, ensuring that calculations based on these values match observable phenomena, such as the periods of satellites and spacecraft around Earth.

semi-major axis

The semi-major axis is one of the most critical parameters in defining an orbit, especially an elliptical orbit. It is the longest radius of an ellipse, extending from its center through one of the foci to the boundary.
In orbit mechanics, the semi-major axis is directly related to the orbit's period via Kepler's Third Law, influencing the gravitational forces at play. A greater semi-major axis generally means a longer orbital period.
For the spacecraft to transition from a circular orbit to an elliptical one, its semi-major axis had to be recalculated, considering the aim of halving the original orbit's period. From the exercise, the new semi-major axis of the elliptical orbit was found using appropriate calculations, ensuring that the distance calculations aligned with achieving rendezvous without collision.