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?

Step by step solution

01

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: $$T^2=\frac{4\pi^2}{GM}\times a^3$$ $$T=\sqrt{\frac{4\pi^2}{GM}\times (2.5\times 10^7)^3}\approx 5790\,\text{seconds}$$

02

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 \(T_{\textrm{ellipse}}=\frac{1}{2}T\approx 2895\) seconds.

03

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 \(a_{\textrm{ellipse}}\) of the elliptical orbit by solving the equation for \(a_{\textrm{ellipse}}\): $$a_{\textrm{ellipse}}^3=\frac{GM}{4\pi^2}\times T_{\textrm{ellipse}}^2$$ $$a_{\textrm{ellipse}}=\left(\frac{GM}{4\pi^2}\times(2895)^2\right)^\frac{1}{3}\approx 2.195\times 10^7\,\text{m}$$

04

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 \(a_{\textrm{ellipse}} - r_{\textrm{min}} = 2.5\times 10^7\,\text{m}\) or \(r_{\textrm{min}} = a_{\textrm{ellipse}} - 2.5\times 10^7\,\text{m}\). Therefore, $$r_{\textrm{min}}\approx 2.195\times 10^7\,\text{m} - 2.5\times 10^7\,\text{m} \approx -3.05\times 10^6\,\text{m}$$

05

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, \(R_{\textrm{min}}\), is the absolute value of \(r_{\textrm{min}}\): $$R_{\textrm{min}}=|-3.05\times 10^6\,\text{m}|=3.05\times10^6\,\text{m}\approx 3050\,\text{km}$$

06

Compare the minimum radius with Earth's radius

Now we can compare the minimum radius of the new elliptical orbit, \(R_{\textrm{min}}\approx 3050\,\text{km}\), with Earth's radius \(6370\,\text{km}\). Since \(R_{\textrm{min}}<6370\,\text{km}\), 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.

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