Questions: The orbital speed (s) of an earth satellite is related to its distance (r) from the earth's center by the formula √r = .2029 × 10^7 / 3. If the satellite's orbital speed is 4 × 10^3 meters per second, find its altitude a (in meters) above the earth's surface, as shown in the illustration. Round the answer to the nearest meter.

Solution Steps

The problem provides the orbital speed \( v \) of the satellite as \( 4 \times 10^3 \) meters per second. The formula relating the orbital speed \( v \) to the distance \( r \) from the Earth's center is: \[ v = \sqrt{\frac{2.029 \times 10^{13}}{r}} \]

Square both sides of the equation to eliminate the square root: \[ v^2 = \frac{2.029 \times 10^{13}}{r} \] \[ r = \frac{2.029 \times 10^{13}}{v^2} \]

Substitute \( v = 4 \times 10^3 \) meters per second into the equation: \[ r = \frac{2.029 \times 10^{13}}{(4 \times 10^3)^2} \] \[ r = \frac{2.029 \times 10^{13}}{16 \times 10^6} \] \[ r = \frac{2.029 \times 10^{13}}{1.6 \times 10^7} \] \[ r = 1.268125 \times 10^6 \text{ meters} \]

The altitude \( a \) is the distance from the Earth's surface to the satellite. Given the Earth's radius \( R \) is approximately \( 6.4 \times 10^6 \) meters: \[ a = r - R \] \[ a = 1.268125 \times 10^7 - 6.4 \times 10^6 \] \[ a = 6.68125 \times 10^6 \text{ meters} \]

Final Answer