Question

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).

Answer 1

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