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` what will be the mean density of the earth if R is its radius and g is the acceleration due to gravity acting on it’s surface?`

11 months ago

If R is the radius of Earth and g is the acceleration due to gravity on the Earth surface then mean density of Earth is

Its value near the surface of the earth is 9.8 ms-2. Therefore, the acceleration due to gravity (g) is given by = GM/r2.Density is defined as Mass / Unit Volume.Taking the globe as a sphere (it’s actually an oblong spheroid) the volume is calculated by: 4/3 pi R^3.The acceleration due to gravity at the surface of the Earth “g” = G M(e) / R^2This gives us the mass of the Earth: M(e) = g R^2 / GUsing definition of Density: g R^2 / G / 4/3 pi R^3 = (3 g) / (4 G pi R)Plugging in book values, we obtain D = 5,500 kg / m^3 which matches most commonly accepted value for density of Earth.

11 months ago

Let's take an object, say it's a ball that you threw upwards then it's weight mg which is the gravitational force acting on it will be equal to Newton's equation for universal gravitation which is F = GMm/d^2, where G is Newton's universal gravitational constant, M is earth's mass, m is the mass of object in free fall, and d distance from the center of the earth, being the radius R here. You can ignore the height it reaches as it is such a tiny increase in the R, and also all things on earth's surface have weights which are given by mg, with the same g, and the distance between any object on earth's surface and the center of earth's mass is the radius of the earth, R with little variation with latitudes. So we can write it mathematically as follows : GMm/R^2 = mg, which gives you g = GM/R^2 which when solved for M gives you M = gR^2/G. (So you can see also that g varies directly with the mass of any planet or moon, and inversely to the second power of the radius, and M varies with g, and the second power of the radius directly). Now we have a mathematical expression for the mass of the earth M = gR^2/G as derived above ; just beautiful mathematics ! As you can see the smaller mass m cancels out and is never a determinant for the g or the M. The earth's volume is given by the volume of a sphere which is 4/3pR^3 where p is pi. To calculate density you have to divide mass by volume. So let's do that here and get an equation for the density of the earth as follows : D = M/V, hence D = (gR^2/G)/4/3pR^3 which gives you the mean density equation : D = (3/4)g/pGR or D = 3g/4pGR. The only variable here is R in the above equation for the earth, with 3g/4pG being a constant for earth, and both g and R for other planets, and moons will vary with different values of g and R. The acceleration due to gravity on earth, g, varies slightly only based on latitudes, as earth is not a perfect sphere, but an oblate spheroid as stated by Newton 331 years ago !!! As you can see if you keep g constant and vary the R or the radius, then greater the radius the lower the density, and smaller the radius the greater the density(with mass remaining the same inherently) showing an inverse relationship of mean density and the radius R. If you keep R constant and vary g, then greater the g, greater is the density, and smaller the g, the lower is the density(showing variable mass within the same volume) in a direct proportionality. It is to be noted that Newton was able to estimate the mass of the earth working from his own creation, the universal law of gravitation and his equation, with an ingenious approach(before G would be experimentally calculated by Cavendish hundred years later) and was able to estimate the mass of the earth off by a very tiny percentage, way less than 0.1 % ; Just sheer genius ! Kaiser T, MD(Life long physics, mathematics, and science enthusiast).

11 months ago

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