Thursday
May 5, 2016

# Homework Help: Phys Science

Posted by 45843 on Tuesday, July 19, 2011 at 4:33pm.

Good afternoon tutors..I need help on a few problems.Im really stuck..

1) the space shuttle orbits at an altitude of approximately 200km above the surface of the earth. If the radius of the earth itself is 6380km, whats the acceleration de to gravity at the altitude of the spae shuttle?
• Phys Science - tchrwill, Thursday, July 21, 2011 at 1:08pm

<<How does the acceleration due to gravity vary over the earth's surface?>>

As we normally think of it, the static value of gravity on, or above the surface of a spherical body is directly proportional to the mass of the body and inversely proportional to the square of the distance from the center of the body and is defined by the expression g = GM/r^2 = µ/r^2 where GM = µ = the gravitational constant of the body (G = the Universal Gravitational Constant and M = the mass of the body) and r = the distance from the center of the body to the point in question.
We normally think of our weight as being the force with which gravity pulls us toward the center of the earth, or the force our bodies exerts on the ground we stand on. One would normally think that their weight is the same regardless of where they are on the earth's surface. In fact, one's weight varies with location on the earth. Why you ask? It is due to the Earth's rotation on its axis which 1) affects the shape and mass distribution of the earth and 2) produces an outward centrifugal force on a body on the surface. Let me "try" to explain.

The Universal Law of Gravitation, F = GMm/r^2, and the derived expression for gravity, g = GM/r^2, assumes that the central body is spherical. G and M are constant by definition and for a sphere, r also becomes constant, resulting in a constant g. Under the influence of gravity alone, the Earth would undoubtedly be a perfect sphere. However, the Earth's rotation on its axis creates equal and opposite centripetal and centrifugal forces on the Earth's mass removed from the rotational axis. The further from the axis or rotation, the greater the forces. The end result is a stretching of the Earth's mass between the axis of rotation and the outer surface, the maximum stretch ocurring at the equator. This stretch, or bulge, results in the Earth's equatorial radius being ~13 miles greater than the polar radius, 3963 miles vs 3950 miles. Thus, the Earth is not a pure sphere but an oblate spheroid producing a measurable difference in the surface radius the further one gets from the poles coupled with more mass being concentrated under the eqator than at the poles.

Due to this difference in radii, the Earth's shape is usually referred to as an oblate spheroid, slightly flattened at the poles with a symmetrical equatorial bulge. The mean equatorial diameter is given as ~7926.77 miles and the mean polar diameter as ~7900 miles. The degree of polar flattening is expressed by the oblateness of the spheroid, defined as f = (equatorial radius - polar radius)/equatorial radius which is currently given as 1/298.257 or ~.00335364 in the latest 1999 Astronomical Almanac.
A great circle of longitude through the poles may be viewed as an ellipse with a major axis, 2a = 7926.77 and a minor axis, 2b = 7900. This results in the distance from the center of the ellipse to a focus of the ellipse being c = sqrt[a^2 - b^2] = 325.45 miles which, in turn, results in an ellipse eccentricity of e = c/a = 325.45/3963.385 = .082115. The polar equation of the Earth's longitudinal great circle ellipse then becomes
r = a(1 - e^2)/[1 + e(cosv) =
3963.385(1 - .082115^2)/[1 + .082115(cosv)] =
3936.661/[1 + .082115(cosv)]
where r is the radial distance from the focus to the surface in miles and cosv is the cosine of the angle between the major axis and the point of interest (the latitude), measured from the end of the ellipse nearest to the reference focus. Thus, when v = 0, r becomes equal to 3637.931 miles (3963.385 - 325.45) and when v = 180º, r becomes equal to 4288.839 (3963.385 + 325.45). When v = 94.71º, r becomes equal to 3963.385, which is equal to the semi-major axis, as it should be from r(94.71º) = sqrt[b^2 + c^2].

This gives a reasonable definition of the Earth's sea level surface anywhere along a longitudinal great circle, or great ellipse. Actual regional Geodetic and Geocentric Coordinates data may be found in the Explanetary Supplement to the Astronomical Almanac, Edited by P. Kenneth Seidelmann of the U.S. Naval Observatory, Washington, DC, University Science Books, 1992, Chapter 15.

You need only determine the mean height above sea level of your town to add to the calculated radius of the Earth at the latltude of your town. You can then calculate the value of gravity at your location.