Question

On November \(27,1963,\) the United States launched Explorer \(18 .\) Its low and high points above the surface of Earth were approximately 119 miles and 123,000 miles (see figure). The center of Earth is the focus of the orbit. Find the polar equation for the orbit and find the distance between the surface of Earth and the satellite when \(\theta=60^{\circ}\). (Assume that the radius of Earth is 4000 miles.)

Short answer

The polar equation for the orbit is \(r = \frac{65,559.5 miles(1 - 0.931^2)}{1+ 0.931\cos(\theta)}\). The distance from the satellite to the surface of Earth when \(\theta = 60^{\circ}\) needs to be calculated.

Step-by-step solution

Calculate the semi-major axis of the ellipse

To find the polar equation of the ellipse, we first need to find the lengths of the major and minor axes of the ellipse. Given that the high point is 123,000 miles and the low point is 119 miles from the Earth's surface, respectively, we add these to the radius of Earth to get the apogee and perigee. The apogee \(a = 123,000 miles + 4000 miles = 127,000 miles\) and the perigee \(p = 119 miles + 4000 miles = 4119 miles\). The semi-major axis \(a\) is the average of the apogee and perigee: \(a = \frac{apogee + perigee}{2} = \frac{127,000 miles + 4119 miles}{2} = 65,559.5 miles\).

Calculate the semi-minor axis of the ellipse

To find the semi-minor axis \(b\), we use the formula \(b = \sqrt{apogee \times perigee} = \sqrt{127,000 miles \times 4119 miles} = 27,980.34 miles\).

Calculate the eccentricity and form the polar equation

We can find the eccentricity \(e\) from the relationship \(e = \frac{\sqrt{a^2 - b^2}}{a}\) which gives \(e = \frac{\sqrt{(65,559.5 miles)^2 - (27,980.34 miles)^2}}{65,559.5 miles}= 0.931\). Given this, the polar equation for the orbit in form \( r = \frac{a(1 - e^2)}{1+ e\cos(\theta)}\) is \(r = \frac{65,559.5 miles(1 - 0.931^2)}{1+ 0.931\cos(\theta)}\).

Find the distance between the surface of Earth and the satellite when \(\theta=60^{\circ}\)

We substitute \(\theta = 60^{\circ}\) into the polar equation, and calculate the result, the distance from the center of Earth. Subtract the radius of the Earth (4000 miles) from this result to get the height of the satellite from the surface of Earth at \(\theta = 60^{\circ}\).