Has the loss of mass by the Sun over the last 4 billion years been enough to affect planetary orbits?


Fron the ISS ,our sun is a dazzling star (Credit: NASA/ISS). The luminosity of the Sun is 200 trillion trillion watts or 2 x 10^33 ergs per second. From Einstein’s famous equation, E = mc^2, and using c = 3 x 10^10 centimeters/sec, the Sun’s luminosity is equal to a loss of mass from the fusion cycle of about 2 x 10^12 grams/second. Over one year this is 7 x 10^19 grams, and over the entire life of the Sun to date is about 3.1 x 10^29 grams. The mass of the Sun is 4 x 10^33 grams so this loss equals 0.008 percent of its current mass. The mass of Jupiter is about 0.1 percent of the Sun’s current mass, so over the Sun’s entire lifetime to date, it has lost barely 0.08 percent of Jupiter’s mass, or about the mass of the Earth.

We can estimate how much this mass loss would have changed the orbit of the Earth by approximating the orbital dynamics as the balance between kinetic and gravitational potential energy or 1/2 mV^2 = GMm/R where m = mass of Earth, and M = mass of Sun. We see that a reduction in the Sun’s mass by a factor of of 0.00008 causes an increase in the Earth-Sun distance if the kinetic energy of the Earth is held constant. This means that over the last 4.5 billion years, we can estimate that the Earth’s orbit has increased by about 0.00008 x 93 million miles or about 7,000 miles; about the Earth’s own diameter!

The Sun also produces a ‘solar wind’ of particles at a rate of about 10^-14 solar masses per year. The NASA illustration shows the general idea of what this wind does as it travels through interplanetary space. In 4 billion years this amounts to about 0.001 percent of the Sun’s mass, which to the level of our approximations is a factor of 8 times smaller that the mass loss from converting some of its mass into light. However, both the solar luminosity and solar wind have not been constant over 4 billion years, with the sun having been fainter long ago, and its wind having been much stronger when it was first born.

The overall effects of these mass loss rates can be significant when dynamicists try to predict the long-term orbits of planets. We know that small changes in any physical parameter in these ‘non-linear’ mathematical theories can produce substantial changes in the locations of planets in their orbits. It would not surprise me if the sun loosing 1 Earth mass over 4 billion years might also have a significant effect in predictions of where planets are in the distant future.

What is the distance from the Sun to the Earth for each month of the year?


We all know that Earth orbits our sun in an elliptical path, which means that for certain times of the year it is closer to the sun than for others. Here is a distorted view of the basic orbit (Credit: Wikipedia), but beware that the scales are all wrong. There is only a 5 million kilometer difference between the longest and shortest lengths of the ellipse!

According to the 1996 US Ephemeris, on the 21st of each month, the distance to the Sun is:

Month...............distance............

January             0.9840       147,200,000
February            0.9888       147,900,000
March               0.9962       149,000,000
April               1.0050       150,300,000
May                 1.0122       151,400,000
June                1.0163       152,000,000
July                1.0161       152,000,000
August              1.0116       151,300,000
September           1.0039       150,200,000
October             0.9954       148,900,000
November            0.9878       147,700,000
December            0.9837       147,200,000

.................................

where the first column gives the distance in Astronomical Units so that 1.0 AU = 149,597,900 kilometers By the way, I have rounded all of the distance numbers to 4 significant figures. If you want a mathematical formula that gives you the distances in AU using these table entries for month number N, it looks like this:

d(N) = 1.0000 + 0.0163 cos [ (2pi/12)*((N-1)/12) + 200*2pi/360]

As you can see from the table, the Earth is farthest from the Sun when the Northern Hemisphere is in the summer season (July 3), and closest in the winter (January 3). If you were living in the Southern Hemisphere, the seasons are reversed so that in the summer, the Sun is closest and in the winter it is farthest.

Why does Venus rotate backwards from the other planets?


The rotation period of Venus cannot be decided through telescopic observations of its surface markings because its featureless thick atmosphere makes this impossible. In the 1960’s, radar pulses were bounced off of Venus while at its closest distance to the Earth, and it was discovered that its rotation period, its day, was 243.09 +/- 0.18 earth days long, but it rotated on its axis in a backwards or retrograde sense from the other planets. The above image was created by NASA’s Magellan spacecraft whos radar imaging technique was able to detect surface feaatures as small as a few kilometers across.

If you were to look down at the plane of the solar system from its ‘north pole’ you would see the planets orbiting the Sun counterclockwise, and rotating on their axis counterclockwise. Except for Venus. Venus would be rotating clockwise as it orbited the Sun counterclockwise. Venus is not alone. The axis of Uranus is inclined so far towards the plane of the solar system that it almost rolls on its side as it orbits the Sun.

What accounts for the extreme inclinations of the rotation axis of Venus and Uranus? For years it was thought that in the case of Venus that Earth was the culprit. It is a curious fact that as Venus rotates three times on its axis in 729.27 days, Earth goes twice around the Sun ( 728.50 days) This has suggested to many astronomers that Earth and Venus are locked into a 3:2 tidal resonance. There are many bodies in the solar system that seem to be locked into various kinds of spin-orbit resonances, especially families of asteroids with the planet Jupiter. Mercury also seems to be gravitationally locked into some kind of resonance with the Sun since its day (58.646 days) and its year ( 87.969 days) are also in the proportion of 3:2.

Forces acting on spinning bodies result in some peculiar acrobatics. For instance, if you take a spinning top and give it a push, it will begin to wobble in a manner called precession. The axis of the Earth makes a 26,000 year wobble with an amplitude of tens of degrees. This is all due to the influence of the Moon’s tidal attraction of the Earth. In the case of Venus, however, the gentle gravitational forces it may receive over billions of years to place it in a 3:2 resonance with the Earth don’t seem to be strong enough to tip the entire planet over to make its rotation retrograde. The best, current, ideas still favor some dramatic event that occurred while Venus (and Uranus for that matter) were being formed. It is known from the cratering evidence we see on a variety of planetary surfaces, that soon after the planets were formed, there were still some might large mini-planets orbiting the Sun. One of these may have collided with the Earth, dredging up material that later solidified into our Moon. The satellites of the outer planets are probably representitives of this ancient population of bodies. Venus may have experienced an encounter with one of these large bodies in which, unlike for Earth, the material didn’t form a separate moon, but was absorbed into the body of Venus. In addition to mass and kinetic energy, this body would also have contributed angular momentum. The result is that the new spin direction and speed for Venus was seriously altered from its initial state, which could have been very Earth-like. Today, the result of that last, ancient collision is Venus with a retrograde rotation.

This theory may also apply to Uranus provided that the collision happened before the 15 satellites themselves were captured or formed. Their orbital planes look very uniform and show no evidence for a dramatic gravitational event such as a collision. It may be, too, that the Uranian collision event dredged up matter and flung it into orbit around Uranus, and out of this were formed the larger, coplanar, moons of Uranus.

This is, clearly, a complicated and not well understood phenomenon. The facts for Venus point towards a collision event to put its axis and rotation in the retrograde sense. The tidal action of the Earth on Venus, acting steadily over billions of years, then established the 3:2 spin-orbit resonance. Every 2 earth years, the exact same portion of the Venerian ( Cytherian) surface faces Earth. Could there be some sub- surface concentration of mass on this portion of Venus that the Earth can grab onto to create the tidal lock?

How long would a trip to Mars take?


Contrary to the ‘point and shoot’ idea, an actual trip to Mars looks very round-a-bout as the figure above shows for a typical ‘minimum cost’ trajectory. This, by the way, is called a Hohman Transfer Orbit, and is the mainstay of interplanetary space travel. All you need to do is give your payload a few km/sec of added speed at Earth to place you into the right elliptical orbit whose aphelion is at the location of mars, and then when you get there you fire your rockets in the opposite direction a second time to reduce your speed by a few km/sec to put you into a mars-capture orbit. You spend your entire flight coasting between the planets with no rockets firing. That’s what makes it a cheap flight in terms of fuel, but an expensive one in terms of flight time. In the end, it is always fuel cost that wins.

The travel time depends on the exact details of the orbit you take between the Earth and Mars. The typical time during Mars’s closest approach to the Earth every 1.6 years is about 260 days. Again, the details depend on the rocket velocity and the closeness of the planets, but 260 days is the number most often estimated, give or take 10 days. Some high-speed transfer orbits could make the trip in as little as 130 days.

That said, chemical rockets are very inefficient for interplanetary travel with specific impulses of about 250 seconds and maximum speeds of 10 km/sec. For some really quick trips ,engineers consider advanced ion propulsion systems with specific impulses of over 5000 seconds and maximum speeds of 100 km/sec and travel times of a month. Nuclear-thermal rockets can do even better with speeds up to 10,000 km/sec. At these speeds, a trip to Mars becomes less than a week!

What is the Typical Temperature on Mars?


The daytime surface temperature is about 80 F during rare summer days, to -200 F at the poles in winter. The air temperature, however, rarely gets much above 32 F.

The temperatures on the two Viking landers, measured at 1.5 meters above the sursurface, range from + 1° F, ( -17.2° C) to -178° F (-107° C). However, the temperature of the surface at the winter polar caps drop to -225° F, (-143° C) while the warmest soil occasionally reaches +81° F (27° C) as estimated from Viking Orbiter Infrared Thermal Mapper.

In 2004, the Spirit rover recorded the warmest temperature around +5 C and the coldest is -15 Celsius in the Guisev Crater.

The Curiosity Rover landed in August 2012 and has been steadily building up an enormous data base of meteorological conditions and measurements. Here from its equatorial location in Gale Crater are its temperature readings. Credit: NASA/JPL-Caltech/CAB(CSIC-INTA)/FMI/Ashima Research. This pair of graphs shows about one-fourth of a Martian year’s record of temperatures (in degrees Celsius) measured by the Rover Environmental Monitoring Station (REMS) on NASA’s Curiosity rover. The data are graphed by sol number (Martian day, starting with Curiosity’s landing day as Sol 0), for a period from mid-August 2012 to late February 2013, corresponding to late winter through the end of spring in Mars’ southern hemisphere.

Why don’t Mercury and Venus have moons?


This colorful view of Mercury was produced by using images from the color base map imaging campaign during MESSENGER’s primary mission. These colors are not what Mercury would look like to the human eye, but rather the colors enhance the chemical, mineralogical, and physical differences between the rocks that make up the planet’s surface. Image credit: NASA / Johns Hopkins University Applied Physics Laboratory / Carnegie Institution of Washington.

Most likely Mercury and Venus have no moons because they are too close to the Sun.

Any moon with too great a distance from these planets would be in an unstable orbit and it would be captured by the Sun. If they were too close to these planets they would be destroyed by tidal gravitational forces. The zones where moons around these planets could be stable over billions of years is probably so narrow that no body was ever captured into orbit, or created in situ, when the planets were first being accreted.

What are the specifics? Lets do the calculation of forces!

Mercury: How far from Mercury would a satellite be at which point the gravity force of the sun equals that of mercury?

G(Mp)m/rs^2 = G(Ms)m/R^2

were M is the masss of the sun, m is the mass of the satellite, rs is the distance of the satellite from the center of mercury and R is the distance from Mercury to the sun. The value for the satellite mass, m, cancils, as does the constant of gravity, G, so for R = 35 million km, Ms = 2×10^30 kg, Mp = 3.3×10^23 kg, we get with some algebra that rs = 14,000 km.

The tidal limit for the satellite, which is the point at which the tidal gravity force of Mercury shatters the satellite, is given by the formula R = 1.26 x Rm (Rho m/Rho M)^1/3 where Rho m is the satellites density and Rho M is the planets density. For a satellite made of mercury material the densities are the same so the tidal radius is 1.26 x Mercury’s radius of 2440 km, or 3070 kilometers.

So the roughly stable orbit range for such a satellite is between 3000 km and 14,000 km from the center of Mercury, or between 600 and 11,000 km from its surface. For Venus, the same calculation woud give a significantly wider orbit range from 7600 km to 162,000 km from the planet’s center, or 1600 to 156,000 km from its surface.

Clearly, Venus is a far more promissing place to stuff one or more moons, so it is a puzzle why it has none, unless the catastrophic impact that tilted Venus’s rotation axis also stripped off any primordial, orbiting moons.

What is a red giant star?


This is the first direct image of a star other than the Sun, made with the Hubble Space Telescope. Called Alpha Orionis, or Betelgeuse, it is a red supergiant star marking the shoulder of the winter constellation Orion the Hunter. The Hubble image reveals a huge ultraviolet atmosphere with a mysterious hot spot on the stellar behemoth’s surface. The enormous bright spot, which is many hundreds times the diameter of Sun, is at least 2, 000 Kelvin degrees hotter than the surface of the star.

A red giant star is a star with a mass like our Sun that is in the last phase of its life. Hydrogen fusion reactions have become less efficient in the core region, and with gravitational collapse of the core, the fusion reactions now occur in a shell surrounding the core. This increases the luminosity of the star enormously (up to 1000 times the Sun) and it expands. The outer layers then cool to only 3000 K or so and you get a red star, but its size is now equal to the orbit of Mercury or Venus…or even the Earth! After a few more millions of years, the star evolves into a white dwarf-planetary nebula system and then it’s all over for the star.

The closest red giant star to our sun is Gamma Crucis (also referred to as Gacrux). It is the third-brightest star in the Southern Cross. Unlike its blue-white neighbors in the constellation, Gacrux is a bright red giant. Gacrux is also considered the nearest red giant to Earth, at a distance of roughly 88 light years.

Can you see stars from the bottom of a well?


From the bottom of a 6-foot diameter well, at a depth of say 50 feet, the sky would subtend the same angle to the eye as a 1.5 inch diameter tube of length 1 foot held up to the eye. This is about the same size as the paper tube in American ‘kitchen towel’ paper. If you do this experiment in the daytime you will see that no stars come out by sighting through such a tube. So, by direct experiment with a similar geometry, the answer is no, you cannot see stars in the daytime at the bottom of a well.

As a caviat, Ken Tapping an astronomer at the NRC Herzberg Institute of Astrophysics notes: “If we were on the Moon’s surface, where there is no atmosphere, we could simply shade out the Sun and the reflected glare from the ground and see the stars perfectly well. On the Earth, our atmosphere tends to scatter the sunlight, which is what makes the sky look blue. This blue is sufficiently bright that it is very difficult to see the stars through it, although on really clear days, in dark places such as the bottom of a well, where reflected light from things on the ground isn’t reaching your eyes, it is sometimes possible to see a star or two.”

Another answer, is given by David Hughes in the Quarterly Journal of the Royal Astron. Soc., 1983, vol. 24, pp 246-257.

This mistaken notion was first mentioned by Aristotle and other ancient sources, and was widely assumed to be correct by many literary sources of the 19th century, and even believed by some astronomers. But every astronomer who has ever tested this by experiment came away convinced it was impossible.

Separate experiments to attempt to see Vega and Pollux through tall chimneys were performed by J. A. Hynek and A. N. Winsor. They were unable to detect the stars under near perfect conditions, even with binoculars. The daytime sky is simply too bright to allow us to see even the brightest stars (although Sirius can sometimes be glimpsed just after the Sun rises if you know exactly where to look.) Venus can be seen as a tiny white speck but again, you have to be looking exactly at the right spot. According to Starwaders.com

Venus is a tiny point of light during daylight and it cannot be seen if the eye is not focussed at the furthest distance, i.e. infinity. In between looking at what is apparently a blank blue sky, briefly flick your gaze every few seconds to an object on the horizon or even onto a tree a hundred meters away. This pulls the eye lens into long distance focus. You could have been looking directly at the point where Venus was and not seen a thing, but having nudged your eyes into infinity focus, be totally surprised to see the bright Venus diamond suddenly become “as clear as daylight”. You will wonder why you could not see it before and why others around you cannot see it.

The most likely explanation for the old legend is that stray bits of rubbish get caught in the updraft and catch the sunlight as they emerge from the chimney. It is possible to see stars in the daytime with a good telescope, as long as it has been prefocused and can be accurately pointed at a target.

What is the relationship between a star’s color and its temperature?


Color and temperature are related by the famous Planck, black body formula shown here in a NASA-Webb Space Telescope illustration. Actually, the formula gives the intensity of a black body at any wavelength given its temperature. The wavelength where the peak of this curve occurs is determined by the temperature of the black body. You then have to relate this peak wavelength of emission to how the human eye identifies color in the visual part of the electromagnetic spectrum. For example, the wavelength of peak emission is

                                      2897
Wavelength (micrometers) = ------------------
Temperature (K)


so that for the Sun with a temperature of 5700 K, the peak of the black body curve occurs at 2897/5700 = 0.51 micrometers or 5100 Angstroms. A cool M-type star can have T = 2500 K so their peak emission occurs at 1.16 micrometers in the deep or ‘far’ red part of the spectrum.

How big is Proxima Centauri?


Proxima Centauri, shown hear in a Hubble image (Credit:NASA), is a dM5e star (dwarf M5 emission-line star) with a luminosity of 0.00006 times the sun that was discovered in 1915 by the Scottish astronomer Robert Innes, the Director of the Union Observatory in South Africa, when it was at that time 0.1 light years closer to our sun than Alpha Centauri.

Because of Proxima Centauri’s proximity to Earth, its angular diameter can be measured directly and is 1.02 ± 0.08 milliarcsec. At its distance, that means it is about one-seventh the physical diameter of the Sun. By comparison, it is about 50% larger that Jupiter.

Today, its orbit around Alpha Centauri (distance = 4.395 light years) now puts it at a distance of 4.223 light years according to the Hipparcos Satellite.

Proxima is located about 13,000 AUs from Alpha Cantauri A and B. The star is located roughly a fifth of a light-year from the AB binary pair and, if gravitationally bound to it, may have an orbital period of around half a million years. According to Anosova et al (1994), however, its motion with respect to the AB pair is hyperbolic.

Alpha Centauri A and B orbit each other at a distance of about 2.2 billion miles (3.6 billion kilometers), a bit more than the distance from the Sun to planet Uranus. It takes 80 years for them to complete an orbit. Proxima Centauri is nearer to Earth than the other two stars, by the rather large distance — roughly 10,000 times the distance from Earth to the Sun. All three known stars in the system were born about 4.85 billion years ago, astronomers believe. Our Sun began shining about 4.6 billion years ago. The A and B stars are both about the same temperature as the Sun. Proxima Centauri is about seven times smaller than the Sun. It contains just enough mass to cause hydrogen to burn, and it is much cooler and, intrinsically, only about 1/150th as bright as the Sun. This small star is barely a star at all, in fact. Its mass is just above that of brown dwarfs, a class of object that seems to straddle the definition between stars and planets. Though 150 times more massive than Jupiter, Proxima Centauri is only about 1.5 times bigger than the planet.

On August 24, 2016, the European Southern Observatory announced the discovery of Proxima b, a planet orbiting the star at a distance of roughly 0.05 AU (7,500,000 km) and an orbital period of approximately 11.2 Earth days. Its estimated mass is at least 1.3 times that of the Earth. The equilibrium temperature of Proxima b is estimated to be within the range where water could exist as liquid on its surface, thus placing it within the habitable zone of Proxima Centauri. Previous searches for orbiting companions had ruled out the presence of brown dwarfs and supermassive planets orbiting Proxima.