Aphelion/Perihelion of Halleys comet

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A.U.) / (75 years)² * ((2/(16 A.U)) - (1/16.8 A.U.))= 1.929 x 10^6 km²/s²Taking the square root of this gives us the speed of Halley's Comet at aphelion and perihelion, which is approximately 1390 km/s.In summary, the orbit of Halley's Comet has a high eccentricity of 0.967 and a period of 76 years. Using the formula for calculating the semimajor axis, the maximum and minimum distances from the sun were determined to be 2.666 x 10^
  • #1
zanazzi78
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Q.Halley's comet is in an elliptic orbit about the sun. The orbit eccentricity is 0.967 and the period is 76 years. Taking the mass of the sun to be [tex] 2 \times 10^30 kg[/tex] abd the usual value of G, determine the max and min distances of the comet from the sun.
Now I've worked out teh answers but they differ from the values I've found on the net,(my guess is the value I've used for the solar mass isn't very accurate!) so would you mind taking a second to have a look at what I've done to see if I'm correct.
A.
using...
[tex]a=( \frac{GM_{\odot}T^2}{4 \Pi ^2})^\frac{1}{3} [/tex]
i got
[tex]a= \sqrt[3]{ \frac{ (6.67 \times 10^-11 )( 2 \times 10^30 )( 2.4 \times 10^9)^2}{4 \Pi^2}}
= 2.01 \times 10^9 m
[/tex]
from
[tex]e= \frac{a-R_{min}}{a}
[/tex]
[tex]
R_{min} = 6.633\times1 0^6
[/tex]
then to get R_max i used
[tex]
R_{max} = 2a - R_{min}
= (2)(2.01\times10^9) - (6.633\times 10^6)
= 2.666\times 10^{16} m
[/tex]
the problem is i`ve found a value for R_min = 8.9x10^10 and R_max = 5.3x10^12 !
 
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  • #2
zanazzi78 said:
[tex]a= \sqrt[3]{ \frac{ (6.67 \times 10^-11 )( 2 \times 10^30 )( 2.4 \times 10^9)^2}{4 \Pi^2}}
= 2.01 \times 10^9 m
[/tex]

Check that calculation again. You have a power (-11 + 30 + 18)/3, it should be in the power 12 range.
 
  • #3
I`m blind to my own ignorance, thank you for pionting out the error.
 
  • #4
I am trying to figure out the speed of Halley's Comet in km/sec. at aphelion and perihelion

I am using

v² = (4Π²a³) / (P²) * ((2/r) - (1/a))

where a = mean distance from the sun (semimajor axis of the ellipse)
P= sidereal period (75 years)
r = distance of the object from the Sun at a given instant

a=16.8 A.U. c=16 A.U.

"r" at perihelion = a-c "r" at aphelion = a + c
 

What is the Aphelion of Halleys comet?

The Aphelion of Halleys comet refers to the point in its orbit around the sun where it is farthest away from the sun. For Halleys comet, this distance is approximately 35 astronomical units (AU) or 35 times the distance between the Earth and the sun.

What is the Perihelion of Halleys comet?

The Perihelion of Halleys comet refers to the point in its orbit around the sun where it is closest to the sun. For Halleys comet, this distance is approximately 0.6 AU or 0.6 times the distance between the Earth and the sun.

How often does Halleys comet reach its Aphelion and Perihelion?

Halleys comet reaches its Aphelion and Perihelion every 75-76 years. This is because it has an elliptical orbit around the sun, with an average orbital period of 75.3 years.

What factors affect the Aphelion and Perihelion of Halleys comet?

The Aphelion and Perihelion of Halleys comet are affected by the gravitational pull of other planets in the solar system, particularly Jupiter. This can slightly alter the comet's orbit and change the distance it reaches from the sun during each orbit.

When will Halleys comet reach its next Aphelion and Perihelion?

Halleys comet is predicted to reach its next Aphelion and Perihelion in 2061. This will be the comet's 31st recorded appearance since its first observed appearance in 240 BC.

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