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photoshop1234 [79]
3 years ago
9

Kepler's third law is founded on a mathematical formula that is based on the inverse relationship between a planet's orbital vel

ocity and its distance from _____. the sun the earth the moon some other entity
Physics
2 answers:
zepelin [54]3 years ago
6 0

Answer:

Kepler's third law is founded on a mathematical formula that is based on the inverse relationship between a planet's orbital velocity and its distance from <u>the Sun</u>.        

Explanation:

According to Kepler's third law of planetary motion, the square of period of orbit of a planet (P) is directly proportional to the cube of distance from the Sun (a).

P^2=\frac{4\pi^2 a^3}{GM}

It was derived from equating gravitation force with centripetal force:

\frac{GMm}{a^2}=\frac{mv^2}{a}

There is inverse relation between orbital velocity (v) and its distance from the sun (r)

Cloud [144]3 years ago
4 0
Kepler's third law is founded on a mathematical formula that is based on
the inverse relationship between a planet's orbital velocity and its distance
from the sun.
You might be interested in
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
onsider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST applies
cluponka [151]
The answer is C) A girl hangs by both hands, motionless, from a trapeze.
5 0
3 years ago
A cricket player catches the ball leaning towards to the ground,why?​
Vadim26 [7]

Answer:

Explanation:

As it’s difficult to catch it from up.

Gravitational force will pull us when we jump.

If gravity was not there, he could catch the ball. But he will float in the sky after that.

That’s the answer

3 0
3 years ago
In order for an object to have kinetic energy it must have a mass and a ?
almond37 [142]

Answer:

Velocity

Explanation:

  • The mechanical energy of the body is defined as the sum of the potential energy and kinetic energy.

                                   E = P.E + K.E

  • The potential energy of a body is due to the height from the surface of the earth.

                                  P.E = mgh

  • The kinetic energy of the is possessed by the body due to the virtue of its motion,

                                  K.E = ½ mv²

  • If there is no velocity associated with the body, there is no K.E in the body.
8 0
3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
2 years ago
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