Solve the problem.
-The orbit of a satellite had a semimajor axis of $a = 6955 \mathrm {~km}$. Calculate the period of the satellite. (Earth's mass $= 5.975 \times 10 ^ { 24 } \mathrm {~kg}$ and $\mathrm { G } = 6.6720 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm {~kg} - 2$ ).
A) 0.18 sec
B) 96.2 min
C) 124 min
D) 9250 hr

Question

Quizplus · Accepted Answer

The period of the satellite can be calculated using Kepler's third law, which relates the period of an orbit to the semi-major axis of the orbit and the mass of the central body. The formula for the period $T$ in seconds is given by $T = 2\pi \sqrt{\frac{a^3}{GM}}$, where $a$ is the semi-major axis in meters, $G$ is the gravitational constant, and $M$ is the mass of the Earth. Converting $a$ to meters gives $a = 6955 \times 10^3$ meters. Plugging in the values, $T = 2\pi \sqrt{\frac{(6955 \times 10^3)^3}{6.6720 \times 10^{-11} \times 5.975 \times 10^{24}}}$ results in a period of approximately 5812 seconds, or 96.87 minutes, which is closest to 96.2 minutes.

Solve the Problem a=6955 kma = 6955 \mathrm {~km}a=6955 km Calculate the Period of the Satellite

Solve the Problem $a = 6955 \mathrm {~km}$ Calculate the Period of the Satellite