Acceleration due to gravity: Difference between revisions

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imported>'Dragon' Dave McKee
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Considering a body with the mass ''M'' as a source of a gravitational field, the strength of that field, or the
Considering a body with the mass ''M'' as a source of a gravitational field, the strength of that field, or the
gravitational acceleration, is given by <math>\vec g = -G \frac{M}{r^2} \frac{\dot{r}}{r}</math>.  
gravitational acceleration, is given by <math>\vec g = -G \frac{M}{r^2} \frac{\vec{r}}{r}</math>.  
The modulus of ''g'' is <math>g = G \frac{M}{r^2}</math>.
The modulus of ''g'' is <math>g = G \frac{M}{r^2}</math>.



Revision as of 21:16, 23 February 2008

Considering a body with the mass M as a source of a gravitational field, the strength of that field, or the gravitational acceleration, is given by . The modulus of g is .

Here G is the gravitational constant, G = 6.67428×10-11 Nm2/kg2, r is the distance between a body of mass m and the center of the gravitational field, is the vector radius of that body having the mass m. If the source of the gravitational field has a spherical shape, then r is the sphere’s radius. Taking into account that the Earth is an oblate spheroid, the distance r is not that of a sphere and varies from the equator to the poles.

OblateSpheroidAngles.png

A normal section (on the equatorial plane) is almost an ellipse, so, r can be done by:


where re and rp are the equatorial radius and polar radius, respectively and θ is the latitude, or the angle made by r with the equatorial plane.

References

1. V.Dorobantu and Simona Pretorian, Physics between fear and respect, Vol. 3, Edited by Politehnica

                                         Timisoara, 2007, ISBN 978-973-625-493-2