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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).

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Answer:

Step-by-step 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
* 10^8 \ km

Therefore,
T^2 = ca^3


c= (365^2)/((1.496 * 10^8)^3)

c = 3.9791
* 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 = (10759.22^2)/(3.9791 * 10^(-20))


a^3 = 2.9092 * 10^(27)


a= \sqrt[3]{2.9092 * 10^(27)}

a = 1.4275
* 10^9 \ km

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

Using the formula


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

G =
6.674 * 10^(-11)

M =
1.989* 10^(30) \ kg

r = perihelion


v ^2= GM ( (2)/(r)-(1)/(a))


(10.18 * 10^3) ^2= 6.674 * 10^(-11) * 1.989 * 10^(30) ( (2)/(r)-(1)/(1.425 * 10^(12)))


7.8068 * 10^(-13)= (2)/(r)-(1)/(1.425 * 10^(12))


(2)/(r) = 1.4824 * 10^(-12)


r = (2)/(1.4824 * 10^(-12))


r = 1.349 * 10^(12)

Similarly, the perihelion is expressed by the equation,

r = a(1 - e)

where ;

e= eccentricity


1.349 * 10^(12) = 1.425 * 10^(12) ( 1 - e)


1.349 * 10^(12) - 1.425 * 10^(12)= - 1.425 * 10^(12) (e)


-7.6* 10^(10)= - 1.425 * 10^(12) (e)


(-7.6* 10^(10))/(- 1.425 * 10^(12))= (e)

e ( eccentricity) = 0.0533

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


r = 1.425 * 10^(12) ( 1 + 0.0533)


r = 1.50 * 10^(12) \ m

to nautical miles, we have:


r = 1.50 * 10^(12) * 0.00054 \ nautical \ mile

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

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

the body of the solar system is Saturn

User Sandeep Chikhale
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