Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet, ![r=4\ AU=1.496\times 10^{11}\ m](https://tex.z-dn.net/?f=r%3D4%5C%20AU%3D1.496%5Ctimes%2010%5E%7B11%7D%5C%20m)
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
![T^2=\dfrac{4\pi^2}{GM}r^3](https://tex.z-dn.net/?f=T%5E2%3D%5Cdfrac%7B4%5Cpi%5E2%7D%7BGM%7Dr%5E3)
M is the mass of the sun
![T^2=\sqrt{9.96\times 10^{14}}\ s](https://tex.z-dn.net/?f=T%5E2%3D%5Csqrt%7B9.96%5Ctimes%2010%5E%7B14%7D%7D%5C%20s)
T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
You find yourself in a place that is unimaginably <u>hot and dense</u>. A r<u>apidly changing</u><u> gravitational field</u><u> </u>randomly warps space and time. Gripped by these huge fluctuations, you notice that there is but a single, unified force governing the universe, you are in the early universe before the Planck time.
<h3>What is Planck time?</h3>
The Planck time is approximately<u> 10^-44 seconds</u>. The smallest time interval, or "zeptosecond," that has so far been measured is <u>10^-21 seconds</u>. A photon traveling at the speed of light would need one Planck time <u>to traverse a distance of one </u><u>Planck length</u>.
<h3>What is Planck length?</h3>
Planck units are a set of measuring units used only in particle physics and physical cosmology. They are defined in terms of <u>four universal </u><u>physical constants</u> in such a way that when expressed in terms of these units, these physical constants have the numerical value 1. These units are a system of natural units because its definition is <u>based on characteristics of nature</u>, more especially the characteristics of free space, rather than a selection of prototype object, as was the case with Max Planck's original 1899 proposal. They are pertinent to the study of unifying theories like quantum gravity.
To learn more about Plank time:
brainly.com/question/23791066
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Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .