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Nitella [24]
3 years ago
7

If the distance between two objects increased, which of the following things could keep the force of gravity between the two obj

ects the same?
a. increasing the mass of one of the objects.
b. decreasing the mass of one of the objects.
c. increasing the velocity of one of the objects.
d. decreasing the velocity of one of the objects.
Physics
1 answer:
AURORKA [14]3 years ago
7 0
The answer A hope this helped

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<h2>Answers:</h2>

<h2>(a) </h2>

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

F=\frac{GMm}{{r}^{2}}     (1)

This expression can also be written as:

F=GMm{r}^{-2}    (2)

If we derive this force F respect to the distance r between the two masses:

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Taking into account GMm are constants:

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

Or

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<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies.  </h2><h2 />

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.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction.  </h2>

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.

<h2>(d) </h2>

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

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

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

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

Isolating -2GMm:

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

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

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

Isolating -2GMm:

-2GMm=X({(10000km)}^{3})     (9)

Making (7)=(9):

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Then isolating X:

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

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

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