" When the apple falls,does the apple also attract the earth?

Physics

1 answer:

Gre4nikov [31]3 years ago

8 0

Answer:

yes

Explanation:

masses attract each other

with a teeny-tiny force

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Which of the following is true for the angular momentum of rotating bicycle wheel

Blababa [14]

<span>Is the following sentence true or false? Newton's first law does object's mass concentration and its axis of rotation increases, its rotational inertia The bicycle wheels at rest have no angular momentum, and the bicycle will fall over easily.</span><span>
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3 years ago

Kepler’s third law can be used to derive the relation between the orbital period, P (measured in days), and the semimajor axis,

NikAS [45]

Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant

Given:
P = 687 days
A = 1.52 AU

Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³

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3 years ago

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$