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<https://everythingaboutspace.quora.com/If-we-had-the-chance-to-actually-know-the-true-size-of-the-universe-how-do-you-think-this-would-have-affected-our-appro>
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43.6667° N    70.2667° W  Founded January 1970
2022-2023: LXI
Sunrise: 6:41 a.m.
Sunset: 5:10 p.m.
Civil twilight begins: 6:12 a.m.
Civil twilight ends: 5:39 p.m.
Sun's host constellation: Capricornus
Moon phase: Waning crescent(39% illuminated)
Moonrise: 1:42 a.m.
Moonset: 11:29 a.m. (2/15/23)
Julian date: 2459986.21

THE BI-WEEKLY ASTRONOMER
Tuesday, February 14, 2023
Pandora Potpourri

Why is the distance between the Sun and Earth not increasing if the space
itself is expanding according to the Big Bang theory?
<https://www.quora.com/Why-is-the-distance-between-the-Sun-and-Earth-not-increasing-if-the-space-itself-is-expanding-according-to-the-Big-Bang-theory>

It is all a matter of scale and distance. Over cosmological scales -i.e
along the scales of galactic clusters- the universal expansion proves to be
a significant factor related to celestial motions. However, on a
comparatively micro-scale such as the solar system, the gravitational force
exerted between bodies easily exceeds the negligible Universal expansion
along that scale. As a rough analogy, the solar system isn’t affected by
the Universal expansion just as a water molecule isn’t torn asunder by an
Earthquake. The electromagnetic forces within the molecule are far more
powerful than the tectonic activity that so profoundly affects continents.

Even galaxies, themselves, are generally bound together strongly enough to
retain their dimensions despite the expansion. As another example, the
Milky Way Galaxy and the Andromeda Galaxy, the nearest major spiral galaxy
to the Milky Way, are currently moving toward each other due to their
mutual gravitational attraction in defiance of the expansion.

[The Universal expansion is easily overpowered by the gravitational
bindings of galaxies and solar systems within those galaxies. Despite its
current 2.2 million light year distance, the Andromeda Galaxy is due to
collide with the Milky Way in approximately four billion years despite the
Universal expansion. Image credit: NASA/JPL]

The solar system is far too small to be affected by something as grand and
far-reaching as the Universal expansion.

How long would it take a spacecraft to travel one light-year?
<https://www.quora.com/How-long-would-it-take-a-spacecraft-to-travel-one-light-year>

Currently, five spacecraft are escaping the solar system:

   - Pioneer 10
   - Pioneer 11
   - Voyager 1
   - Voyager 2
   - New Horizons

Provided that none of them crashes into any celestial object -a highly
remote contingency- each of these craft will eventually travel one light
year. (In fact, they will all likely travel much farther over time.) How
long will each one of them require to move such a distance?

[AU = astronomical unit, equal to Earth’s average distance from the Sun]

   - Pioneer 10. Speed relative to the Sun: 2.51 AU/year. Current distance
   from Sun: 132.9 AU. Will be one light year away in 24,143 years [Launched
   3 March 1972]
   - Pioneer 11 Speed relative to the Sun: 2.36 AU/year. Current distance
   from Sun: 110.9 AU. Will be one light year away in 26,750 years [Launched
   6 April 1973]
   - Voyager 1 Speed relative to the Sun: 3.57 AU/year. Current distance
   from Sun: 159 AU. Will be one light year away in 17,670 years [Launched
   5 September 1977]
   - Voyager 2 Speed relative to the Sun: 3.23 AU/year. Current distance
   from Sun: 132.5 AU. Will be one light year away in 19,538 years [Launched
   20 August 1977]
   - New Horizons Speed relative to the Sun: 2.90 AU/year. Current distance
   from Sun: 55.3 AU. Will be one light year away in 21,788 years [Launched
   19 January 2006]

The graphic above shows the different flight paths of the five spacecraft
mentioned above. [Image credit: Heavens-Above.com]

I wanted to restrict my answers to those spacecraft that will actually
traverse such a long distance. The Parker Solar Probe will attain the
highest velocity of any spacecraft when it makes a close approach to the
Sun in 2025. During this close approach, it will move at 690,000 km/hr
(430,000 miles/hour). Were it able to maintain this speed while moving away
from the Sun -it won’t- it would require only 1,563 years to travel one
light year.


Why does the Moon rotate around the Earth, whereas the Sun's gravity is
greater than the Earth’s?
<https://www.quora.com/Why-does-the-Moon-rotate-around-the-Earth-whereas-the-Suns-gravity-is-greater-than-the-Earth-s>

The moon revolves around Earth because the former is well within the
latter’s Hill Sphere. Any object within Earth’s Hill Sphere will be
dominated gravitationally by our planet. In fact, every celestial body is
surrounded by this sphere, the extent of which is determined principally by
its mass and its distance from its parent body.*

[Image credit: ScienceABC]

The above image shows the moon within the boundary defined as Earth’s Hill
Sphere. However, the image is not to scale because Earth’s Hill Sphere
extends 1.47 million kilometers above our planet’s surface. (Refer to chart
below.) Since the moon’s average distance equals about 0.384 million
kilometers, it is well within Earth’s Hill Sphere. Even though the moon is
slowly moving away from Earth at the glacial rate of 3.8 cm/year, it will never
venture outside of this sphere. If it did, however, it would ‘detach’ from
Earth and establish its own orbit around the Sun.

[Image credit: JPL Development Ephemeris]

The chart above shows the Hill Sphere radii for the solar system’s major
bodies. Note: Neptune’s Hill Sphere is the largest of them because of
Neptune’s greater distance from the Sun. Pluto’s HS is smaller because it
is far less massive than Neptune.

[Note: Yes, a moon can in theory have a moon of its own provided that the
smaller moon is within the larger moon’s Hill Sphere, but outside the Roche
Limit, the boundary at which the moon would be torn apart by the larger
moon’s tidal forces.]

*The Hill Sphere of a body of mass m when in orbit around a larger body of
mass M can be approximated by the following formula:

Hill sphere = a(1-e) (m/3M)^1/3

a = the smaller body’s semi-major axis

e = the smaller body’s orbital eccentricity




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