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Kepler's third law can be used to derive the relation between the orbital period, [tex]$P$[/tex] (measured in days), and the semimajor axis, [tex]$A$[/tex] (measured in AU) of an orbiting body. The relation is given by the equation [tex]$P^2 = k A^3$[/tex], where [tex][tex]$k$[/tex][/tex] is a constant value for all bodies orbiting that star. The semimajor axis of Mars is 1.52 AU, and its orbital period is about 687 days. What is the value of the constant [tex]$k$[/tex]?

A. [tex]$4.52 \times 10^2$[/tex]

B. [tex][tex]$7.44 \times 10^{-6}$[/tex][/tex]

C. [tex]$1.34 \times 10^5$[/tex]

D. [tex]$2.21 \times 10^{-3}$[/tex]



Answer :

To determine the value of the constant [tex]\( k \)[/tex] using Kepler's third law, we start with the known values for the semimajor axis [tex]\( A \)[/tex] and the orbital period [tex]\( P \)[/tex] of Mars. According to Kepler's third law, the relationship between the orbital period ([tex]\( P \)[/tex]) and the semimajor axis ([tex]\( A \)[/tex]) is given by the equation:

[tex]\[ P^2 = k \cdot A^3 \][/tex]

Given:
[tex]\[ A = 1.52 \, \text{AU} \][/tex]
[tex]\[ P = 687 \, \text{days} \][/tex]

We need to solve for [tex]\( k \)[/tex]. Rearrange the equation to solve for [tex]\( k \)[/tex]:

[tex]\[ k = \frac{P^2}{A^3} \][/tex]

First, calculate [tex]\( P^2 \)[/tex]:

[tex]\[ P^2 = 687^2 = 471969 \][/tex]

Next, calculate [tex]\( A^3 \)[/tex]:

[tex]\[ A^3 = (1.52)^3 \approx 3.518208 \][/tex]

Now, we can find [tex]\( k \)[/tex]:

[tex]\[ k = \frac{471969}{3.518208} \approx 134394.875 \][/tex]

After calculating, we need to match this value with the provided options:
A. [tex]\( 4.52 \times 10^2 \)[/tex]
B. [tex]\( 7.44 \times 10^{-6} \)[/tex]
C. [tex]\( 1.34 \times 10^5 \)[/tex]
D. [tex]\( 2.21 \times 10^{-3} \)[/tex]

We compare our calculated value [tex]\( 134394.875 \)[/tex] with the given options. The option that closely matches [tex]\( 134394.875 \)[/tex] is:

C. [tex]\( 1.34 \times 10^5 \)[/tex]

Therefore, the value of the constant [tex]\( k \)[/tex] for Mars is:

[tex]\[ \boxed{1.34 \times 10^5} \][/tex]

Hence, the correct answer is option C.

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