Physics, 26.11.2019 06:31 williamnason123

The moon and earth rotate about their common center of mass, which is located about 4700 km from the center of earth. (this is 1690 km below the surface.)

a. calculate the magnitude of the acceleration due to the moon's gravity at that point.
b. calculate the magnitude of the centripetal acceleration of the center of earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). comment on whether or not they are equal and why they should or should not be.

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Answer from: twocutebabe26

a) $a_c=3.41x10^{-5} \frac{m}{s^2}$

b) $a_c=3.34x10^{-5}\frac{m}{s^2}$

If we analyze the two values obtained for the centripetal acceleration we see that are similar. Based on this we can say that the centripetal force would be similar to the gravitational force between the Earth and Moon.

Explanation:

1) Notation and important concepts

Centripetal acceleration is defined as "The acceleration experienced while in uniform circular motion. It always points toward the center of rotation and is perpendicular to the linear velocity."

Angular frequency is defined as "(ω), or radial frequency, measures angular displacement per unit time and the units are usually degrees (or radians) per second. "

T = 27.3 d represent the time required by the Earth around an specific point given in the problem

G= universal constant $6.673x10^{-11}\frac{Nm^2}{kg^2}$

M= represent the mass of the Moon=7.35x10^{22}kg

2) Part a

$a=\frac{GM}{r^2}$ (1)

The Earth-Moon Distance Is 3.84x10^8 km (average Value) and both rotates at a point located about 4700 km from the center of Earth, so the radius for this case would be the difference between these two values

r=(3.84x10^8 km)-(4.7x10^6)=3.793x10^8 m

Since we have the radius now we can replace into equation (1)

$a=\frac{(6.673x10^{-11}\frac{Nm^2}{kg^2})(7.35x10^22 kg)}{(3.793x10^8 m)^2}=3.41x10^{-5} \frac{m}{s^2}$

And the acceleration due to the Moon's gravity would be $3.41x10^{-5} \frac{m}{s^2}$ at the point required.

3)Part b

For this case we can find the centripetal acceleration from this formula:

$a_c =r \omega^2$ (2)

But on this case we don't have the angular frequency so we can find it with this formula

$\omega =\frac{2\pi}{T}$ (3)

But since the period is on days we need to convert that into seconds

$27.3dx\frac{86400s}{1d}=2358720sec$

Replacing the value into equation (3) we got:

$\omega =\frac{2\pi}{2358720}=2.664x10^{-6}\frac{rad}{s}$

Now we can find the centripetal acceleration with the equation (2), the new radius on this case since our reference is the Earth and the point is located 4700km=4700000m from the center of Earth then the new value for the radius would be r=4700000m

$a_c =r \omega^2 =(4700000m)(2.664x10^{-6}\frac{rad}{s})^2=3.34x10^{-5}\frac{m}{s^2}$

Answer from: Quest

Velocity=total displacement/timefrom house to store is 500 from store to neighbor is 200 therefore total displacement=500+200=700metrestherefore,700metres/910seconds=0.8metres per second

Answer from: Quest

Physical science b final exam bernoulli’smovepowercompound translucent the sun i was having such a pain for this searching every question so i hope i made someone happy

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