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velikii [3]
2 years ago
13

Applying newton's version of kepler's third law (or the orbital velocity law) to the a star orbiting 40,000 light-years from the

center of the milky way galaxy allows us to determine ______.
Physics
1 answer:
Ugo [173]2 years ago
5 0

Applying Newtons version of Kepler's third law or the orbital velocity law to the star orbiting 40000 light years from the center of the Milky Way Galaxy allows us to determine the mass of the Milky Way Galaxy that lies within 40000 light years in the galactic center.

<h3></h3><h3>What is orbital velocity law?</h3>

The orbital velocity law states that, the orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius of the body. Earths orbital velocity near its surface is around 8km/sec if the air resistance is disregarded.

In space exploration, orbital velocity is a crucial topic. Space authorities heavily rely on it to comprehend how to launch satellites. It aids scientists in figuring out the velocities at which satellites must orbit a planet or other celestial body to prevent collapsing into it. The speed at which one body orbits the other body is known as the orbital velocity. The term "orbit" refers to an object's consistent circular motion around the Earth. The distance between the object and the earth's centre determines the orbit's velocity.

To know more about orbital velocity law, refer brainly.com/question/11353717

#SPJ4

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Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
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Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

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(Correct Answer is above)

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