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3 years ago

13

Question: Why do both the moon and the earth pull on the spaceship the entire time? Why must you be closer to the moon before th

e pull of the earth and moon are equal?

1 answer:

Ira Lisetskai [31]3 years ago

8 0

-- There's no limit to the distance of gravitational forces.
There's gravitational force between Pluto and the lint in your
pocket ... not much, but it's there, and it can be calculated.

So there's ALWAYS gravitational force between the Earth and the
spaceship, AND ALSO between the Moon and the spaceship.
Even before it's ever launched !

-- The Earth has about 80 times as much mass as the Moon has,
so you have to be much closer to the Moon before the gravitational
forces in each direction are equal.

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Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

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If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

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$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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2) liquid-solid is the most common type of solution.

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