LOCAL ASTRONOMY – A FEW QUESTIONS AND ANSWERS
A note by Eamon Henry. Date: 23 September 2002.

Introduction:
Recently I asked myself a few questions on astronomical matters within the Solar System. Not only did I not know the answers, but I could find no answers within easy reach, either. So I decided to do my own sums. I share with you the answers. I hope you enjoy them!

A few constants etc. are needed. Force = mass * acceleration, where * means “multiplied by”. Powers of 10 such as 10 to the 7th power (meaning 7 tens multiplied together) will be written below as 10(7). One divided by 10(7) will be written 10(-7) . The constant “pye” (3.141592654) will be written pye. R to the power of 2 will be written Rsquared or R*R. (This unusual notation is to fit blog text conditions, which do not accept superscripts for powers of numbers or of symbols.)
The unit of force is the “Newton”, which gives to a mass of one kilogram (kg) an acceleration of one metre (m) per second, per second. So all units will be expressed in kg and m, with lots of powers of 10 as well.
At a long distance, massive bodies act with gravitational force upon each other as if all the mass were concentrated at the centre of gravity of each such body.
We need G, the universal constant of gravitation, of value 6.67259*10(-11)

The basic fact is that a body in orbit has a centrifugal force on it per kg mass (to make it fly away) given by V*V /R, where R is its distance from the “centre of force”, and V is its velocity in metres per second. It is kept from flying away by the force of gravity between it and the body it is orbiting, given by G* M1* M2 /( R*R) , here R being the distance in metres between the centers of the two bodies and M1 and M2 their masses in kg. So we have two formulae to calculate what we want, which is a useful check, as will appear below. We keep life simple by assuming circular orbits, of length 2*pye* R, which is alright because many planetary orbits are ellipses nearly circular (a circle is an ellipse with its two foci together).

Given values of mass and distance for Sun, Earth, Moon, Mars, Venus, and Jupiter (mostly in Britannica 2002 but some in Patrick Moore’s “Exploring the Night Sky with Binoculars”), we can now ask and answer a few questions:

What gravitational force does the Sun exert on the Earth, on average?
What gravitational force does the Earth exert on the Moon, on average?
How do these forces compare?
What gravitational forces of Sun and Moon affect the sea tides on Earth (diameter 12756 kilometres, same as 7926 miles)?
What force of gravity do Venus, Mars, Jupiter exert on the Earth, when they are at their nearest, approximately? How does this compare with the Moon-Earth force?
Calculate the parameters of a typical artificial satellite orbiting the Earth only a few hundred kilometers up.
Calculate the parameters for a satellite to rotate at same speed as Earth, so as to stay above a fixed point of the Equator (not to complicate our problem unduly).



Masses and Distances and Velocities:

SUN: mass: 1.99* 10(30) kg; mean distance 149.59285* 10(9) metres from centre of Earth

EARTH: mass 5.976* 10(24) kg; equatorial radius 6378,000 metres.

MOON: mass 73.506* 10(21) kg ( 1/ 81.3 of Earth); mean distance 386.0637* 10(6)
metres from centre of Earth.
VENUS: mass 4.87* 10(24) kg; shortest dist. from Earth’s centre 42* 10(9) metres

MARS: mass 6.418* 10(23) kg; shortest dist from Earth’s centre 78.4036*10(9) metres

JUPITER: mass 18.99* 10(26) kg; shortest dist from Earth’s centre 628.4036* 10(9) metres

Shortest distances have been generally approximated as the difference between their average distances from the Sun, as their orbital planes are fairly close together. Venus: mean distance from Sun 108* 10(9) metres.
Mars: mean distance from Sun 228* 10(9) metres
Jupiter: mean distance from Sun: 778* 10(9) metres.
Average velocity of Earth 29784 metres / second on assumed circular orbit of radius average distance from Sun and for 365.25 days period (of one complete circuit)
Average velocity of Moon 1036 metres / second on assumed circular orbit, of period 27.1 days (during which Moon completes 360 degrees of rotation around Earth).

Gravitational Force Results:

Sun on Earth : 35.44* 10(21) newtons ,using V*V /d formula (centrifugal formula)
35.14* 10(21) newtons, using G*M1*M2/(R*R) formula (gravity formula)

Earth on Moon: 20.44*10(19) newtons, using centrifugal formula
19.66*10(19) newtons, using gravity formula.
Thus, the Sun-Earth force is about 179 times (3514 / 19.66) as great as the
Earth-Moon force. A value of about 173 is given by 3544/20.44.

For the other three planets, I have used the ratio of their gravity formulas to that of
Sun-Earth to obtain as follows for approximate shortest distance effect:
Venus 0.56 ( about ½ ) of 1 percent of Earth-Moon force
Mars 0.02 ( about 1/50) of 1 percent of Earth-Moon force
Jupiter almost (0.98 of) 1 percent of Earth-Moon force.
We may notice how relatively small these last three planetary forces are, with Jupiter’s largest being hardly 1 percent of the Earth-Moon force. If these three planets vanished overnight, we would suffer no noticeable changes in our present orbital behavior.


Ocean Tides on Earth:

The Moon pulls on the Earth with an equal and opposite force to that estimated above,
19.66*10(19) newtons, per gravity formula. We may divide this by Earth’s mass 5.976*10(24) kg, and obtain 3.290*10(-5) equal to 0.0000329 newtons as average Moon gravity-force on each kg of Earth’s material.

Ocean tides on Earth are due to variation in the forces of gravity of both Sun and Moon across the diameter of Earth, to be calculated by the inverse distance squared on nearest surface point versus that at opposite surface point of a diameter. G and the two masses are the same, but the distance changes.

So, for the Moon, we compare 1/(386.064 less 6.378)squared with 1/(386.064 plus 6.378)squared, and get 0.000006937 compared with 0.000006493, the first being 6.84 percent larger than the latter. This means, relative to the Earth’s centre, a force some 3.42% larger on the surface point nearest the Moon, matched by a force 3.42% smaller on the surface farthest behind. This causes the tidal bulge towards the Moon at its side, matched by a tidal bulge “falling behind” at the back. These differences are in Moon gravity-force units.

A similar exercise for the Sun gives a ratio difference (from unity) of only 0.0001728 Sun units across the Earth’s diameter. But when we multiply by 179, to get it in Moon units, we obtain 3.1 percent difference, meaning a force 1.55% (in Moon units) more (than at the Earth’s centre) on the face towards the Sun, matched by a force 1.55% less on the opposite face away from the Sun. Without the distorting effects due to gravity pulls of Moon and Sun, the tidal waters would stay as parts of a perfect sphere about the Earth’s centre, for Earth here assumed to be approximately a perfect sphere.

I need only say that the outcome is the sum of percentages 3.42 and 1.55. Thus, Sun and Moon together give a +5.0 % on the nearest facing surface at new moon, matched by a –5.0% at the point farthest behind. A week or so later the Moon is pulling sideways, relative to the Sun, and we have low (neap) tides. At full moon, the effect (and tides) are again about the same as at new moon. We may see how each “falling behind”, relative to the Earth’s centre, is in effect the same as a negative force pulling in the opposite direction. So Sun and Moon do not tend to cancel out their tidal effects at full moon, although they are on opposite sides of the Earth. The fact is that tidal maximum is much the same. However (Moore page 9), Earth is about 147 million km from the Sun in December, but farther away (152 million km) in June. So this winter nearness to the Sun does cause higher tides in winter.

Satellites:
We equate the two formulae, with gravity force on left equal to satellite mass multiplied by centrifugal acceleration on the right, Msat being satellite mass and Mearth being Earth’s mass:

G* Msat* Mearth /(R*R) = Msat*V*V / R
Here R is the distance of the satellite (on circular orbit) from Earth’s centre.
This reduces to (per kg of satellite mass)
R*V*V = G * Mearth, and the right-side product has constant value 39.8754* 10(13) .
So we have to find values of R and V to satisfy this equation.

1) A near-Earth satellite
Given that the Earth’s radius is about 6400 km, allow 400 km as satellite height above the Earth (assumed here to be a perfect sphere) and thus take 6800 km as satellite distance R from Earth’s centre, which is 6.8 * 10(6) metres. This gives V*V as 58.6402909*10(6), so V is 7657.7 metres (about 4.8 miles) per second. And the orbit length , 2* pye* R is 42.72566 million metres. For the given value of V, it takes 5579 seconds to make one circuit, which is 92 minutes and 59 seconds, roughly an hour and a half, in good agreement with what we have found to happen in the real-world experience. And this satellite supposedly passes over at about 400 km above Earth’s surface.

2) A satellite to stay above a fixed point of Earth’s equator
This means the satellite has a period of about 23hours 56 minutes (86160 seconds), as it makes one complete circuit of its circular orbit in the same time as one rotation of Earth.
So we have 2*pye*R /V = 86160, which gives R / V = 13712.79
And again the other equation gives R*V*V = 39.8754* 10(13).
This leads to V*V*V = 29.078982* 10(9), giving V as 3075.105 metres per second.
And R is found to be 4.2168248 * 10(7) metres, which is 42168.25 km (26,203 miles) above Earth’s centre, roughly 22,000 miles above the Earth’s surface.