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Kaylis [27]
3 years ago
13

By using kepler's 3rd law we find that ___________.

Physics
2 answers:
tangare [24]3 years ago
8 0

Answer;

By using kepler's 3rd law we find that;

-A year on Earth is shorter than a year on Saturn.

Explanation;

-Kepler’s 3rd law states that the square of a planet’s orbital period is proportional to the cube of its average distance from the Sun (semi-major axis), which tells us that more distant planets move more slowly in their orbits.

-In other words, if you square the 'year' of each planet, and divide it by the cube of its distance to the Sun, you get the same number, for all planets. The law captures the relationship between the distance of planets from the Sun, and their orbital periods.

maw [93]3 years ago
3 0

By using Kepler's 3rd law we find that A year on Earth is shorter than a year on Saturn (d)

<h3>Further explanation</h3>

Kepler's law is based on the orbits of planets around the sun.

There are 3 laws issued by Kepler

  • 1. Kepler Law 1

the inner planet's orbit around the sun is based on an elliptical path and the sun is the center

  • 2. Kepler Law 2

Imaginary lines are drawn from planets to the sun cover the same area at the same time interval.

  • 3. Kepler Law 3

The square of the orbital period of a planet is proportional to the cube of its average distance to the sun

General formula:

\large{\boxed{\bold{\frac{T_1^2}{r_1^3} =\frac{T_2^2}{r_2^3} }}}

There are choices of questions that need to be completed from the questions above:

a. A day on Earth is shorter than a day on Saturn.

b. A year on Earth is longer than a year on Saturn.

c. A day on Earth is longer than a day on Saturn.

d. A year on Earth is shorter than a year on Saturn.

If we use Kepler 3's Law to see the answer choices then:

  • 1. the length of days of a planet depends on the rotation of each planet

Kepler 3's law is based only on the period of the planet, in this case, the planet's revolution against the sun.

So the determination of the length of the day can not be determined using Kepler 3 Law

So answers a and c do not meet the area of ​​Kepler 3 law

  • 2. Long or short years of a planet based on the planet's revolution against the sun

One year is counted one revolution = one cycle = one orbit

The longer the planet's distance from the sun, the longer it will get one rotation so that the longer the year from the planet

The planets that circle the sun in sequence from the closest are: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune

From this sequence shows that the distance of the planet Saturn is further than that of the planet Earth so

A year on Earth is shorter than a year on Saturn (d)

<h3>Learn more</h3>

The Sun has more gravity than the other plans in our solar system

brainly.com/question/3252750

weight and mass used synonymously on Earth

brainly.com/question/5956881

the Earth has more gravitation pull than the moon

brainly.com/question/1513912

Keywords: Kepler law 3, sun, revolution, earth, Saturn, rotation, year, day

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What is the power of 10 when 0.00503 is written in scientific notation?
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a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde
Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

  • The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal
  • The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

Volume \ of \ a \ cone = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot h

  • Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

  • \dfrac{dV}{dr} = \dfrac{d}{dr}  \left(\dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot  s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

\underline{\left  \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}

\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}

Learn more here:

brainly.com/question/14466080

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