The International Space Station is orbiting at an altitude of about 370 km above the earth's surface. The mass of the earth is 5.97 × 1024 kg, the radius of the earth is 6.38 × 106 m, and G = 6.67 × 10-11 N • m2/kg2. Assuming a circular orbit, (a) what is the period of the International Space Station's orbit? (b) what is the speed of the International Space Station in its orbit?

A. 5521.86 secs

B. 7680.65 m/s

Explanation:

Parameters given:

Radius of orbit of the space station, R = 6380000 + 370000 = 6750000 m

Mass of earth = 5.97 * 10^24 kg

Gravitational constant, G = 6.67 * 10^(-11) Nm²/kg²

A. Orbital period, T, can be obtained using the formula:

T²/R³ = (4 * pi²) / (G * M)

T²/(6750000³) = (4 * 3.142²) / (6.67 * 10^(-11) * 5.97 * 10^24)

T² = (4 * 3.142² * 6750000³) / (6.67 * 10^(-11) * 5.97 * 10^24)

T² = 30490944.39

T = 5521.86 secs

B. Orbital velocity, v, can be obtained by using the formula:

v² = (G * M) / R

v² = (6.67 * 10^(-11) * 5.97 * 10^(-24)) / 6750000

v² = 58992384

v = 7680.65 m/s

double the distance = divide strength by 4

c. magnitude and direction