UnitMath Examples: ( About Earth )

Last Modified: 2 /16 / 2000

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Find the amount of air in Earth's atmosphere.

We know that the standard pressure at sea level is one atmosphere. If we multiply one atmosphere by the surface area of the Earth we will find the total force of the air on the Earth. Then we can apply the formula Force = mass * acceleration to find the total mass of the air. So the mass of the air is ( Earth's Surface Area ) * atm / ( acceleration of gravity ).

`
EarthsRadius: (6371.315 pm 0.437) km;
SurfaceAreaOfEarth: 4 pi EarthsRadius^2 as sq mile; ~ 1.97 e8 sq mile
airMass: atm * SurfaceAreaOfEarth / ag; ~ 5.27 e18 kg
people: 6e9;
airMass / people as MegaTons ~ 0.97 MegaTons `

With the additional fact that the average density of air at sea level is 1.18 g/liter, we can calculate the volume of the air ( if it were all at sea level ).

`
airVolume: airMass / ( 1.18 g/liter ) as cu mile; ~ 1.1 e9 cu mile
( airVolume / people )^(1/3) as miles ~ 0.56 miles
`

So there is 1.2e19 lbs of air around Earth; which comes down to 0.97 MegaTons per person. If All that air were at sea level ( 1 atm pressure ) then there would be1.1e9 cu miles of air or each persons air would fit in a cube with sides of 0.56 miles.

There's a lot of air out there.

The following information is often useful for calculations relating to the Earth. The data is formated as a UnitMath variables, so it can be directly imported to UnitMath.

EarthRadius: ( 6317.315 pm 0.437 ) km;

EarthRadiusEquatorial: ( 6378.533 pm 0.437 ) km;

EarthRadiusPolar: ( 6356.912 pm 0.437 ) km;

EarthMass: ( 5.979 pm 0.004 )e24 kg;

EarthAphelion: ( 1.5207 pm 0.0007 )e8 km;

EarthPerhelion: ( 1.4707 pm 0.0007 )e8 km;

Find geosynchronous orbit. That is the distance from Earth where a satellite will stay in orbit over one spot on Earths equator.

The following equations are used in this calculation: "` Fg = m MassEarth G/r^2 & Fc = m velocity^2/r`". In any stable circular orbit Fg = Fc so: `MassEarth G/r^2 = velocity^2/r`. To be in geosynchronous orbit it must make just one revolution a day; which, is `2 pi radiusGeo / 24 hours`.

The following calculation finds the radius that meets these requirements to the nearest mile.

`
EarthsMass: (5.979 pm 0.004 ) e24 kg;
radiusGeo: 4000 mile to 1/4 MegaMile as mile;
velocity: 2 pi radiusGeo/day;
solve( EarthsMass G/radiusGeo^2 = velocity^2/radiusGeo; radiusGeo: mile ) ~ radiusGeo: ( 26,247. to 26,263. ) mile;
`

So a satellite orbiting at ( 26,247. to 26,263. ) miles from the center of the Earth will be in geosynchronous orbit. That is ( 22,287. to 22,305. ) miles above the surface of the Earth at the equator.

Find the area of the Earth that can be seen from geosynchronous orbit.

The surface area visible is a slice of the Earth; which, is a Zone and Segment of One Base. The formula for the surface area is

` surfaceArea: 2 pi radius * h;`

Where ( see the following diagram for details )

radius: (6371.315 pm 0.437) km;

angle: asin ( radius / radiusGeo );

h: radius ( 1 - sin( angle ) );

The following calculation finds the surface area visible from geosychronous orbit.

`
surfaceArea: 2 pi radius * h;
`

`
radius: (6371.315 pm 0.437) km as mile; ~ 3,959. mile
radiusGeo: ( 26,247. to 26,263. ) mile;
`

`
angle: asin ( radius / radiusGeo ) as deg; ~ 8.7 deg
h: radius ( 1 - sin( angle ) ) as miles; ~ 3,362. miles
`

`
surfaceArea as sq miles; ~ 8.4e7 sq miles
surfaceArea as sq km; ~ 2.2e8 sq km
`

`
Earth'sSurfaceArea: 4 pi sq radius as sq miles; ~ 2.0 e8 sq miles`

`
surfaceArea / Earth'sSurfaceArea as % ~ 42.5 %
`

So a satellite in geosynchronous orbit can see ( and be seen by ) 42.5 % of the Earth at one time. The highest latitude that can be seen is ( 90 - angle ) = 81.3 degrees ( as John E. Prussing noted ), so a geogeosynchronous satellite can see ( and be seen ) from well above the Artic Circle to a good distance into Antaarctica.