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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2 where G is the gravitational constant and r is the distance between the bodies.

(a) Find dF/dr. dF dr =

What is the meaning of dF/dr?

1) dF/dr represents the rate of change of the mass with respect to the distance between the bodies.
2) dF/dr represents the rate of change of the distance between the bodies with respect to the force.
3) dF/dr represents the amount of force per distance.
4) dF/dr represents the rate of change of the mass with respect to the force.
5) dF/dr represents the rate of change of the force with respect to the distance between the bodies.

What does the minus sign indicate?

1) The minus sign indicates that as the distance between the bodies increases, the magnitude of the force decreases.
2) The minus sign indicates that as the distance between the bodies increases, the magnitude of the force increases.
3) The minus sign indicates that the bodies are being forced in the negative direction.
4) The minus sign indicates that the force between the bodies is decreasing.
5) The minus sign indicates that as the distance between the bodies decreases, the magnitude of the force remains constant.

(b) Suppose it is known that the earth attracts an object with a force that decreases at the rate of 4 N/km when r = 20,000 km. How fast does this force change when r = 10,000 km? N/km

Respuesta :

cryssatemp

Answers:

(a)

According to Newton's Law of Gravitation, the Gravity Force is:

[tex]F=\frac{GMm}{{r}^{2}}[/tex]     (1)

This expression can also be written as:

[tex]F=GMm{r}^{-2}[/tex]    (2)

If we derive this force [tex]F[/tex] respect to the distance [tex]r[/tex] between the two masses:

[tex]\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr[/tex]     (3)

Taking into account [tex]GMm[/tex] are constants:

[tex]\frac{dF}{dr}dFdr=-2GMm{r}^{-3}[/tex]     (4)

Or

[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex]     (5)

(b) dF/dr represents the rate of change of the force with respect to the distance between the bodies.  

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

(c) The minus sign indicates that the bodies are being forced in the negative direction.  

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

(d)

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit [tex]N/km[/tex]:

[tex]\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}[/tex]     (5)

We have a force that decreases with a rate 1 [tex]\frac{dF_{1}}{dr}dFdr=4N/km[/tex] when [tex]r=20000km[/tex]:

[tex]4N/km=-2\frac{GMm}{{(20000km)}^{3}}[/tex]     (6)

Isolating [tex]-2GMm[/tex]:

[tex]-2GMm=(4N/km)({(20000km)}^{3})[/tex]     (7)

In addition, we have another force that decreases with a rate 2 [tex]\frac{dF_{2}}{dr}dFdr=X[/tex] when [tex]r=10000km[/tex]:

[tex]XN/km=-2\frac{GMm}{{(10000km)}^{3}}[/tex]     (8)

Isolating [tex]-2GMm[/tex]:

[tex]-2GMm=X({(10000km)}^{3})[/tex]     (9)

Making (7)=(9):

[tex](4N/km)({(20000km)}^{3})=X({(10000km)}^{3}[/tex]       (10)

Then isolating [tex]X[/tex]:

[tex]X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}[/tex]  

Solving and taking into account the units, we finally have:

[tex]X=-32N/km[/tex]>>>>This is how fast this force changes when [tex]r=10000 km[/tex]

mmartinez0901

Answer:a) -2GmM / r^3

b) dF/dr represents the rate of change of the force with respect to the distance between the bodies

c) The minus sign indicates that as the distance between the bodies increases, the magnitude of the force decreases

d) 4 N/km, r=20,000, and r=10,000

-32 N/km

Explanation:

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