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Licemer1 [7]
3 years ago
11

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).
Business
1 answer:
anastassius [24]3 years ago
5 0

Answer:

Explanation:

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

T^2 \alpha  a^3

where;

T = period

a = semi major axis

T = 356 days (for earth)

a = 1 AU = 1.496 \times 10^8 \ km

Therefore, T^2 = ca^3

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

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

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

∴

T^2 = ca^3

a3 = \dfrac{10759.22^2}{3.9791 \times 10^{-20}}

a^3 = 2.9092 \times 10^{27}

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

a = 1.4275 \times 10^9 \ km

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

Using the formula

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

G = 6.674 \times 10^{-11}

M = 1.989\times 10^{30} \ kg

r = perihelion

v ^2= GM ( \dfrac{2}{r}-\dfrac{1}{a})

(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}})

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

\dfrac{2}{r} = 1.4824 \times 10^{-12}

r = \dfrac{2}{1.4824 \times 10^{-12}}

r = 1.349 \times 10^{12}

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity

∴

1.349 \times 10^{12} = 1.425 \times 10^{12} ( 1 - e)

1.349 \times 10^{12}  -  1.425 \times 10^{12}= -  1.425 \times 10^{12} (e)

-7.6\times 10^{10}= -  1.425 \times 10^{12} (e)

\dfrac{-7.6\times 10^{10}}{-  1.425 \times 10^{12}}=  (e)

e ( eccentricity) = 0.0533

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

r = 1.425 \times 10^{12} ( 1 + 0.0533)

r = 1.50 \times 10^{12} \ m

to nautical miles, we have:

r = 1.50 \times 10^{12} \times 0.00054  \ nautical \ mile

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

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

the body of the solar system is Saturn

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