﻿ What are the Gravitational Forces Required to Keep the Planets in their Orbits?
San José State University

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Thayer Watkins
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 What are the Gravitational Forces Required to Keep the Planets in their Orbits?

## The Balance of Gravitational Attraction of the Sun and the Centrifugal Force on the Planet Due to the Curvature of its Orbit

The gravitational attraction on a planet of mass m due to the mass M of the Sun at a distance R is

#### F = GmM/R²

where G is the gravitational constant.

The centripetal acceleration a experienced by a body moving with a tangential velocity of v in a circular orbit of radius is

#### a = v²/R

The force required to hold it in that orbit is

#### F = ma = mv²/R

If T is the orbit period then the tangential velocity is v=(2πR)/T.

The force holding a planet in orbit is a balance of the centrigugal force by the gravitational force and hence

#### F = GmM/R² = mv²/R

Thus there are two ways of computing the force. Here are the results of such computations.

The mass of the Sun is 1.989×1030 kg. The gravitational constant is 6.67384×10-11 m³/(kg*s²)

Data for the Planets and the Computed Level of Force Holding Them in their Orbits
Planet Mass
(kg)
Orbit
(109 m)
Orbit
Period
(Earth days)
Tangential
Velocity
(m/s)
Centrifugal
Force
(Newtons)
Gravitational
Force
(Newtons)
Ratio
Mercury 3.30E+23 58 88 47930 1.3071E+22 1.30217E+22 0.996231843
Venus 4.87E+24 108 224.7 34953 5.50907E+22 5.54233E+22 1.006036955
Earth 5.97E+24 150 365.2 29869 3.55088E+22 3.52211E+22 0.991897547
Mars 6.42E+23 228 687 24135 1.64017E+21 1.63937E+21 0.999511945
Jupiter 1.90E+27 778 4332 13060 4.1657E+23 4.16682E+23 1.000269809
Saturn 5.68E+26 1,429 10760 9658 3.70755E+22 3.69228E+22 0.995879774
Uranus 8.68E+25 2,871 30700 6801 1.39832E+21 1.39786E+21 0.99966901
Neptune 1.02E+26 4,504 60200 5441 6.70405E+20 6.67443E+20 0.995581671

The computed forces should be equal but due to small differences in the manner of computation they are slightly different

Having computed the force in newtons it is now desirable to express that force as something familiar. The force on the Earth is about 3.53×1022 newtons. A newton is the weight on Earth of 102 grams of matter. The force of the Earth due to the gravitational attraction of the Sun is equal to the weight on Earth of about 3.6×1021 kilograms of matter, say water. Water has a density of 1 gram per cubic centimeter, or 1000 kg/m³. Thus 3.6×1021 corresponds to 3.6×1018 cubic meters or 3.6×109 km*sup3; of water. The volume of water in all of Earth's ocean is 5.136×108 km³. Thus the force on the Earth due to the gravitational pull of the Sun is about 7 times the weight of all the water in Earth's oceans.

## Further Bits of Analysis

In the equation

#### GmM/R² = mv²/R

the mass m of a planet can be cancelled out and v may be replaced by 2πR/T leaving

and hence

#### T² = (4π²/(GM))R³

This is Kepler's Law. .