Solve the problem.
-The orbit of a satellite had a semimajor axis of a $= 8872 \mathrm {~km}$. 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) 78.2 min
B) 1.98 hr
C) 12 hr
D) 138.6 min

Question

Quizplus · Accepted Answer

Using Kepler's Third Law, we can find the period T of the satellite in seconds by the formula
$$T^2 = \frac{4\pi^2a^3}{GM}$$
where G is the gravitational constant, and M is the mass of the central body (Earth in this case). Substituting the given values, we get
$$T^2 = \frac{4\pi^2(8872000)^3}{6.6720\times10^{-11}(5.975\times10^{24})}$$
which simplifies to
$$T^2 \approx 5.345\times10^{11}$$
Taking the square root, we get
$$T \approx 2.313\times10^5 \mathrm{~s}$$
which is equivalent to about 3,860 minutes or 64.3 hours. Therefore, the closest choice is D) 138.6 min, which is approximately equal to the actual period.

Solve the Problem =8872 km= 8872 \mathrm {~km}=8872 km Calculate the Period of the Satellite

Solve the Problem $= 8872 \mathrm {~km}$ Calculate the Period of the Satellite