Question

(a) Based on Kepler’s laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h.

(b) What is unreasonable about this result?

(c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit?

Short answer

1. The orbital radius of the Earth satellite is 5090 km.
2. This result is not reasonable.
3. The premise of a 1.00 h orbital period is not possible.

Step-by-step solution

Definition of Kepler’s law

Planets move in elliptical orbits around the sun, according to the first law. The second claim is that a planet's radius vector sweeps out the same amount of area in the same amount of time.

The third law relates the orbital periods of the planets to their distances from the sun.

Calculating orbital radius.

Kepler's third law, written in mathematical form in, relates the period, or time, for one orbit to the radius of the orbit.

$\frac{{T_1}^2}{{T_2}^2}=\frac{{r_1}^3}{{r_2}^3}$

For the Moon, we'll use the subscript "m ", and for the satellite, we'll use the subscript "s".

• The Orbital radius of the moon from the Table 6.2, $r_m=3.84\times {10}^5\text{\hspace{0.33em}km}=3.84\times {10}^8\text{\hspace{0.33em}m}$.
• Time taken by the moon to complete one orbit is, $T_m=0.07481\text{\hspace{0.33em}y}$.
• The Orbital radius of the earth satellite, ${\text{r}}_{\text{s}}\text{=?}$.
• The Orbital period of the earth satellite,$T_s=1\text{\hspace{0.33em}hr}=1\text{\hspace{0.33em}hr}\times \frac{1\text{\hspace{0.33em}d}}{24\text{\hspace{0.33em}hr}}\times \frac{1\text{\hspace{0.33em}y}}{365\text{\hspace{0.33em}d}}=1.14\times {10}^{-4}\text{\hspace{0.33em}y}$

(a)

Applying Kepler’s third law

$\begin{array}{c}\frac{{T_s}^2}{{T_m}^2}=\frac{{r_s}^3}{{r_m}^3}\\ r_s={\left(\frac{T_s}{T_m}\right)}^{\frac{2}{3}}\times r_m\end{array}$

Putting the values ${\text{T}}_{\text{s}}{\text{,\hspace{0.17em}T}}_{\text{m}}{\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}r}}_{\text{m}}$.

$\begin{array}{c}{r}_s={\left(\frac{T_s}{T_m}\right)}^{\frac{2}{3}}\times r_m\\ ={\left(\frac{1.14\times {10}^{-4}y}{0.07481y}\right)}^{\frac{2}{3}}\times 3.84\times {10}^5\text{\hspace{0.33em}km}\\ r_s=5090\text{\hspace{0.33em}km}\end{array}$

Hence, the orbital radius of the Earth satellite is 5090 km.

Determining Unreasonable result

(b)

The earth's radius is 6376 kilometers, whereas the satellite's radius is 5090 km, which is smaller than the earth's radius, implying that the satellite would have to circle within the earth, which is not conceivable. As a result, this finding is not credible.

Determining premise is unreasonable or inconsistent

(c)

The premise of a 1.00 h orbital period is not possible for a satellite because, in this case, the radius of the orbit comes out to 5090 km which is less than the radius of the earth, which is not possible. So, to make it consistent, you have to increase the orbital period.