The gravitational force due to a large body is given by Fcr2

Solution

this can help u by putting the values in da same method.. lets say that you have scalar function f(x,y,z) (scalar meaning that when evaluated you get single number (it can be imaginary thou)). Example f(x,y,z) = 3x^2+x-y^2+z^2*x*y f(1,2,3) = 3*1^2+1-2^2+3^2*1*2 = 18. Now we would like to know that in which direction our function will grow fastest from that poin (x = 1, y = 2, z = 3). For this purpose is gradient: grad(f(x,y,z)) = (df/dx, df/dy, df/dz) = i*df/fx + j*df/dy + k*df/dz This means that gradient is vector and it points to direction where our function grows fastest (when taking differential step). Example if f(x,y,z) is our example function: grad(f(x,y,z)) = [6*x+1+z^2*y, -2*y+z^2*x, 2*z*x*y] Now if we want to know in what direction our function grows fastest at point (x = 1, y = 2, z = 3) we will evaluate our gradien function and we will get a vector: grad(f(1,2,3)) = (25,5,12) We can also calculate how \"steep\" the raise is by taking length of this vector (in cathesis coordinates it is just sqrt(x^2+y^2+z^2) -> | (25,5,12) | = sqrt(25^2+5^2+12^2) = 28.1780 Usually gradient function is visualized when using only two variables, x and y. Now we can think that x and y gives as a point in a map and gradient at that point gives the direction of steepest slope. Example: If our \"hills\" (height map) are generated z = x^2+y^3+cos(x)*y^2*6+600 our gradient will be: [2*x-6*sin(x)*y^2, 3*y^2+12*cos(x)*y] We can generate our scene in 3D: http://img367.imageshack.us/img367/1331/… And if we now calculate the direction of fastest growth (x = -0.1065, y = 5.005) we will get: (15.76, 134.86). As you can see from the picture, this is the right direction. \"yes, but how do u find the gradient function of say: y = -2x^3+24x-32?\" In that case gradient funcion is your derivative function (it can be calculated by same formulas as for surfaces...) so for that function your gradient function is diff(-2x^3+24x-32) = -6*x^2+24