A body in the solar system has a period of 10,759.22 days and a perihelion speed of 10.18 km/s. a. Calculate the aphelion radius in nautical miles (14 points). b. Calculate the eccentricity (2 points). c. Extra credit: Name the object. (6 points).

Respuesta :

Answer:

Explanation:

From the information given, by applying Kepler's 3rd law,

[tex]T^2 \alpha a^3[/tex]

where;

T = period

a = semi major axis

T = 356 days (for earth)

a = 1 AU = 1.496 [tex]\times 10^8 \ km[/tex]

Therefore, [tex]T^2 = ca^3[/tex]

[tex]c= \dfrac{365^2}{(1.496 \times 10^8)^3}[/tex]

c = 3.9791 [tex]\times 10^{20} \ day^2/km^3[/tex]

However, if the body in the solar system has a period of 10.759.22 days, then, a =?

[tex]T^2 = ca^3[/tex]

[tex]a3 = \dfrac{10759.22^2}{3.9791 \times 10^{-20}}[/tex]

[tex]a^3 = 2.9092 \times 10^{27}[/tex]

[tex]a= \sqrt[3]{2.9092 \times 10^{27}}[/tex]

a = 1.4275 [tex]\times 10^9 \ km[/tex]

However, the velocity for a perihelion = 10.18 km/s

Using the formula

[tex]v = \sqrt{GM ( \dfrac{2}{r}-\dfrac{1}{a})}[/tex] to calculate the radius, we have:

G = [tex]6.674 \times 10^{-11}[/tex]

M = [tex]1.989\times 10^{30} \ kg[/tex]

r = perihelion

[tex]v ^2= GM ( \dfrac{2}{r}-\dfrac{1}{a})[/tex]

[tex](10.18 \times 10^3) ^2= 6.674 \times 10^{-11} \times 1.989 \times 10^{30} ( \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}})[/tex]

[tex]7.8068 \times 10^{-13}= \dfrac{2}{r}-\dfrac{1}{1.425 \times 10^{12}}[/tex]

[tex]\dfrac{2}{r} = 1.4824 \times 10^{-12}[/tex]

[tex]r = \dfrac{2}{1.4824 \times 10^{-12}}[/tex]

[tex]r = 1.349 \times 10^{12}[/tex]

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity

[tex]1.349 \times 10^{12} = 1.425 \times 10^{12} ( 1 - e)[/tex]

[tex]1.349 \times 10^{12} - 1.425 \times 10^{12}= - 1.425 \times 10^{12} (e)[/tex]

[tex]-7.6\times 10^{10}= - 1.425 \times 10^{12} (e)[/tex]

[tex]\dfrac{-7.6\times 10^{10}}{- 1.425 \times 10^{12}}= (e)[/tex]

e ( eccentricity) = 0.0533

Aphelion radius in natural miles, r = a( 1+ e)

[tex]r = 1.425 \times 10^{12} ( 1 + 0.0533)[/tex]

[tex]r = 1.50 \times 10^{12} \ m[/tex]

to nautical miles, we have:

[tex]r = 1.50 \times 10^{12} \times 0.00054 \ nautical \ mile[/tex]

radius of aphelion [tex]\mathbf{r = 8.10 \times 10^8}[/tex] nautical miles

In respect to the value of a( i.e [tex]1.4275 \times 10^9 \ km)[/tex]

the body of the solar system is Saturn