Question

A)Some comets are in highly elongated orbits that come very close to the sun at perihelion. The distance from one such comet to the center of the sun is 6.00×10^9m at perihelion and 3.00×10^12m at aphelion. For this comet's orbit, find the semi-major axis.B)In 2017 astronomers discovered a planet orbiting the star HATS-43. The orbit of the planet around HATS-43 has semi-major axis 7.41×109m, eccentricity 0.173, and period 4.39 days. Find the distance between HATS-43 and the planet at its closest approach.Express your answer with the appropriate units

Answer 1

Answer:

A) 1.50x10¹² m

B) 6.13x10⁹ m

Explanation:

A) The semi-major axis is the average of the perihelion and the aphelion distances:  

$SMA = \frac{P + A}{2}$

where SMA: is the semi-major axis, P: is the perihelion distance and A: is the aphelion distance

$SMA = \frac{6.00 \cdot 10^{9} m + 3.00 \cdot 10^{12} m}{2} = 1.50 \cdot 10^{12} m$

So, the semi-major axis for the comet's orbit is 1.50x10¹² m.

B) To find the distance between the star and the planet at its closest approach, that is to say, at the perihelion, we need to use the Kepler's Law of orbits:

$P = SMA (1 - e)$

where e: is the eccentricity

$P = 7.41 \cdot 10^{9} m (1 - 0.173) = 6.13 \cdot 10 ^{9} m$

Therefore, the distance from the planet to the star HATS-43 is 6.13x10⁹ m at perihelion.  

I hope it helps you!