Is Gravity less at the Equator?

In a recent repeat, in the UK, of "Big Stuff" on Channel 5 (8 pm, 28th July 2002) David Leach, the Facilities Manager for Ariane 5 said:

"We're so close to the equator that gravity's less for a start so you're going to get more return for your thrust. So it's easier; you've got the earth actually helping you launch whereas the further north you go .... the gravity increases".
Is gravity less at the equator, and if so, how much less? And is that the reason then, why rockets are usually launched from sites as near to the equator as possible?

If the earth were perfectly spherical then gravity would be the same over the entire planet's surface. Moving away from the surface (and from the centre of mass or centre of gravity - located at the geometrical centre of the earth) results in a reduction of gravitational force given by the inverse square law:

The earth is not a perfect sphere though - apart from lumps and bumps (mountains and valleys) - the tropics bulge out and the polar radius is some 13 miles less than the equatorial one (3973 compared with 3986 miles or 6357 and 6378 km). So at the equator you are 13 miles further from the centre of mass. How much is gravity reduced?

The ratio of equatorial to polar gravity is: (3973/3986)2 = 0.99348, which means that gravity at the equator is about 0.6% less at the equator.(See Museum of Hoaxes for a bit more about this).

Well if that were the reason that Ariane is launched from the equator, it seems an awful lot of effort and expense to save a little bit of rocket fuel, or get a few extra pounds of payload into space. I doubt that the satellites or rockets are actually manufactured in Guiana so a lot of fuel and effort must be used shipping them there, not to mention the launch facilities and personnel.

The Real Reasons Why an Equatorial Launch Site is Preferred

  1. Many satellites are launched into geostationary orbit - this means that they appear to be fixed with respect to earth - in reality they have an orbital period of 24 hours and must also orbit above the equator. Launching from the equator puts a satellite straight into an equatorial orbit. Launching from any other latitude puts a satellite into an orbit inclined to the equator (so it won't be geostationary) and extra rocket fuel is required to adjust the orbit to an equatorial one.

  2. The earth is rotating once in 24 hours - at the equator this amounts to a speed of 1670 km.hr-1 so even when a satellite is on the ground at the equator it's moving easterly at 1670 km.hr-1 or 0.463 km.s-1. The speed for orbit (500 km above the earth's surface) is 7.59 km.s-1. That's why satellites are almost always lanched towards the east. Moving away from the equator reduces this free speed - at the poles it is zero and in between if falls off with the cosine of the launch site's latitude. Thus launching a satellite towards the east from a site on the equator means that the earth's rotation contributes just over 6% to its final velocity. This is equally useful for interplanetary probes - which will also gain from leaving their orbit around the earth in the same direction as the earth revolves round the sun (by picking up the earth's orbital velocity).

Both the Ariane launch site and Cape Canaveral are on the Atlantic coast and usually launch towards the east - this allows the boosters, debris and so on to fall into the ocean harmlessly; also, in the case of US manned flights (when they were equipped with launch escape rockets - Mercury, Gemini and Apollo) the capsules could splash-down in the Atlantic safely in an emergency (they were not designed for a hard landing). The Vandenberg launch site on the west coast of the US is used for polar-orbiting satellites which can be launched in a southerly direction over the Pacific - no advantage is obtained from the earth's rotation for these orbits so the northerly launch site is not a problem.


Notes

According to the Science Data Book edited by R M Tennent, published by Oliver and Boyd (1971):

g (ms-2) = 9.80616 - 0.025928 cos (2L) + 0.000069 cos² (2L) - 0.000003h

where L = latitude
h = height above sea level (in metres)

 

Last updated: 14 June 2007;   © Lawrence Mayes, 2002-07