Question

Suppose that F is an inverse square force field below, where c is a constant.

Answer 1

Answer:

Forces in our Universe

Step-by-step explanation:

a)

First of all we have,

$F(r) = \frac{cr}{|r|^{3} }$

and,

$r = xi+yj+zk$

We need to define a function that allows us to find said change based on r, so one of the functions that shows that change is,

$f(r) = - c /|r|$

That is,

$\nabla f = F$

For this case F is a conservative field and the line integral is independent of the path. Thus, defining  $P_{1} = (x_{1}, y_{1}, z_{1})$ and $P_{2} = (x_{2}, y_{2}, z_{2})$ . So the amount of work on the movement of the object from P1 to P2 is,

$W=\int\limits_c {F} \, dr$

$W= f(P_{1}-P_{2})$

$W= \frac{c}{(x_{2}^2+ y_{2}^2+ z_{2}^2)^{1/2} } -\frac{c}{(x_{1}^2+ y_{1}^2+ z_{1}^2)^{1/2}}$

$W= c(\frac{1}{d1}-\frac{1}{d2} )$

2) The gravitational force field is given by,

$F(r) =-\frac{mMGr}{ |r|^3}$

The maximum distance from the earth to the sun is $1.52*10 ^ 8$ km and the minimum distance is $1.47*10 ^ 8$km. The mass values of the bodies are given by m = $5.97*10 ^ {24}$kg, M = $1.99 *19 ^ {30}$kg and the constant G is $6.67 * 10 ^{ -11 }$ $\frac{Nm ^ 2}{kg^2}$

In this way we raise the problem like this,

$c= -mMG$

$W= -mMG (\frac{1}{1.52*10 ^ 8} -\frac{1}{1.47*10 ^ 8} )$

$W= -(5.95*10^{24})(1.99*10^{30})(6.67*10^{-11})(-2.2377*10^{-10})$

$W \approx 1.77*10^{35}}J$