Answer:
increase
decrease
Explanation:
using formula
Vt=mg/6πηr
so if m increases V increases
r is the denominator so if r increases V decreases
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.
We're given two angular speeds, and we need to solve for a time.
Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .
Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.
So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !
At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:
They're in line, SOMEwhere on the circles, when
(a fraction of one orbit) = (the same fraction of the other orbit)
AND
the total elapsed time is a common multiple of their periods.
Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.
The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.
K (2π/ω seconds) = (K+1) (π/ω seconds)
2Kπ/ω = Kπ/ω + π/ω
Subtract Kπ/ω : Kπ/ω = π/ω
Multiply by ω/π : K = 1
(Now I have a feeling that I have just finished re-inventing the wheel.)
And there we have it:
In the time it takes the slower planet to revolve once,
the faster planet revolves twice, and catches up with it.
It will be 2π/ω seconds before the planets line up again.
When they do, they are again in the same position as shown
in the drawing.
To describe it another way . . .
When Kanye has completed its first revolution ...
Bieber has made it halfway around.
Bieber is crawling the rest of the way to the starting point while ...
Kanye is doing another complete revolution.
Kanye laps Bieber just as they both reach the starting point ...
Bieber for the first time, Kanye for the second time.
You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.
You've given the answer, right there in your question.
The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.
We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about
(1 / √0.66) = 1.23 times
the Earth's radius, so the height is about 910 miles above the surface.
To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.
The tension in the vertical plane will be equivalent to the centripetal force therefore
Here,
m = mass
v = Velocity
r = Radius
The tension in the horizontal plane will be subject to the action of the weight, therefore
Matching both expressions with respect to the tension we will have to
Then we have that,
Rearranging to find the velocity we have that
The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is
Replacing we have that
Therefore the speed of each seat is 4.492m/s
