Question

A new extrasolar planet (outside our solar system) has been discovered. This planet orbits a star of mass 1.03×10³¹ kg with a period of 6.52 years . It is also known that the planet has an eccentricity of 0.694.

Part A

Part complete

What is the semi-major axis of the planet's orbit?

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Semi−majoraxis= 

9.03×10¹¹

  m 

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 Correct

Part B

What is the average velocity of the planet in its orbit

Part C

What is the distance of closest approach to its parent start (perihelion)?

Part D

What is the farthest distance the planet travels away from its parent start (aphelion)?

Answer 1

A. In accordance with Kepler's third law, the square the time period is proportional to the cube of length of semi major axis. Mathematically, it can be expressed as follows: Use Kepler's third law, we get the semi major axes of the planet GM Here, T is the period, G is the universal constant, M is the mass of the star, and a is the semi major axis 12 ( 4π (6.52 years) 365x24x3600 s 1 year (6.67×10-11 N·m2/kg9(1.03x 1031 kg) a 9.027 x10 m Round off the result to 3 significant digits a= 9.03 x 1011 m

B. The average velocity of the planet on its orbit is, The velocity of the planet in perihelion position is, GM(1+e The velocity of the planet in aphelion position is, GM (1-e Hence, the above equation changes as follows GM1+e GM (1+e 1/2 1/2 al (6.67 x10-11 N.㎡/kg,)(1.03x 1031 kg)((1+0.694)1/2f1-0.694)1.2 l 9.027 x101 m 1-0.694 1+0.694 = 3.83x 104 m/s

(c)The formula for the eccentricity of the planet is, The semi major axes of the planet is a= r + r = 2a Hence, the above equation changes as follows, 2 Add the equations (1) and (2), we get 2石=2a(1+e) = (9.03×101 m) (1+0.694) -1.5296x1012 m

Hence, the distance of closest approach to its parent start is, 2(9.027x101 m)-1.5296x1012 m -2.757 x101 m Round off the result to 3 significant digits 2.76x10" m (H)The farthest distance the planet travels away from its parent start (aphelion) is, r1.529 x102 m Round off the result to 3 significant digits