Solve the problem.
-A space shuttle is in a circular orbit $250 \mathrm {~km}$ above the Earth's surface. Use Kepler's third law (with $\mathrm { a } =$ Earth's radius $+ 250 \mathrm {~km} )$ to find the orbital period of the satellite. $\left( \mathrm { G } = 6.67 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm {~kg} ^ { - 2 } \right.$, $\mathrm { M } = 5.97 \times 10 ^ { 24 } \mathrm {~kg}$ and Earth's radius is $6.37 \times 10 ^ { 6 } \mathrm {~m}$ ).
A) T = 90 min
B) T = 9 min
C) T = 1 hr
D) T = 2 hr

Question

Quizplus · Accepted Answer

The orbital period $T$ can be found using the formula derived from Kepler's third law for circular orbits: $T = 2\pi \sqrt{\frac{a^3}{GM}}$, where $a$ is the semi-major axis (Earth's radius + altitude of the orbit), $G$ is the gravitational constant, and $M$ is the mass of the Earth. Substituting $a = 6.37 \times 10^6 \,m + 250 \times 10^3 \,m$, $G = 6.67 \times 10^{-11} \,Nm^2/kg^2$, and $M = 5.97 \times 10^{24} \,kg$, and solving for $T$, we find that the orbital period is approximately 90 minutes.

Solve the Problem 250 km250 \mathrm {~km}250 km Above the Earth's Surface a=\mathrm { a } =a=

Solve the Problem $250 \mathrm {~km}$ Above the Earth's Surface $\mathrm { a } =$