Solve the problem.
-The orbit of a satellite had a period of $\mathrm { T } = 93.11$ minutes. Calculate the semimajor axis of the orbit. (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) $6805 \mathrm {~km}$
B) $4.440 \times 10 ^ { 4 } \mathrm {~km}$
C) $6.805 \times 10 ^ { 6 } \mathrm {~km}$
D) $444 \mathrm {~km}$

Question

Quizplus · Accepted Answer

The semimajor axis of the orbit can be calculated using Kepler's third law, which relates the orbital period $T$ to the semimajor axis $a$ of the orbit around a central body, in this case, Earth. The formula, when solved for $a$, is given by $a = \left( \frac{G M T^2}{4 \pi^2} \right)^{1/3}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $T$ is the orbital period in seconds. Converting $T = 93.11$ minutes to seconds gives $T = 93.11 \times 60$ seconds. Plugging in the values for $G$, $M$, and $T$, and solving for $a$ yields a value in meters, which when converted to kilometers, gives a semimajor axis close to $6805$ km.

Solve the Problem T=93.11\mathrm { T } = 93.11T=93.11 Minutes Calculate the Semimajor Axis of the Orbit

Solve the Problem $\mathrm { T } = 93.11$ Minutes Calculate the Semimajor Axis of the Orbit