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Bess [88]
3 years ago
10

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

Mathematics
1 answer:
Andre45 [30]3 years ago
5 0

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 ^ 8km. 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

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