ganeshie8
  • ganeshie8
Figure shows three uniform spherical planets that are identical in size and mass. The periods of rotation T for the planets are given, and six lettered points are indicated—three points are on the equators of the planets and three points are on the north poles. Rank the points according to the value of the free-fall acceleration g at them, greatest first.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
|dw:1447477221400:dw|
Astrophysics
  • Astrophysics
Fc=Fg
Astrophysics
  • Astrophysics
Well the gravity at b, d, and f would be less than it is at the equator right

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Astrophysics
  • Astrophysics
Sorry would be stronger
ganeshie8
  • ganeshie8
Since they are ideal spheres, I think the gravity should be same everywhere on the surface ?
Astrophysics
  • Astrophysics
That's a good point
Kainui
  • Kainui
Yeah, I don't see why rotation would affect \[F=G\frac{mM}{R^2}\] So based on that, the force should be the same at all 6 points... But maybe there's something I'm not including? Is this something to do with the Coriolis effect? I have no clue lol.
ganeshie8
  • ganeshie8
there is also centrifugal force on the surface if we switch to rotational frame
Astrophysics
  • Astrophysics
Yes! For that purpose I was thinking it maybe b, d, and f would have a stronger gravity but I barely remember first year physics haha...crap
ganeshie8
  • ganeshie8
what is coriolis effect ?
Kainui
  • Kainui
So the free fall acceleration at the equator is larger because as you fall you don't move but the earth below your feet does. So then that means I think: e>c>a>b=d=f
Astrophysics
  • Astrophysics
I don't think coriolis effect has anything to do with it
ganeshie8
  • ganeshie8
does that mean you weigh more at poles and less at equator if i get it wrong, it would be the opposite of above...
Astrophysics
  • Astrophysics
Well, it means the points at equator are moving faster than points at the pole
ganeshie8
  • ganeshie8
thats coriolis effect ?
Astrophysics
  • Astrophysics
Think so man, I barely remember this stuff
ganeshie8
  • ganeshie8
equator region moving faster than the pole region
ganeshie8
  • ganeshie8
makes sense... i have just seen a short video on why hurricanes rotate as we speak
Kainui
  • Kainui
This is sorta what I'm thinking, but I don't know what the Coriolis effect is so I dunno lol. |dw:1447480177688:dw|
ganeshie8
  • ganeshie8
so coriolis effect can be ignored here as there is no movement parallel to longitudes here ?
Kainui
  • Kainui
No I have no idea what the coriolis effect is, so I have no clue
Astrophysics
  • Astrophysics
I think we can say that, and why'd you say it then kai!
Kainui
  • Kainui
Lol most of my knowledge consists of knowing that things exist and then looking them up and trying to understand them just well enough to solve my problem, and if it's interesting then I'll learn more lol.
ganeshie8
  • ganeshie8
shouldn't the length of arrows be "decreasing" as a planet with shorter period provides a greater centripetal force ? |dw:1447480471830:dw|
Kainui
  • Kainui
Oh right I was backwards I was thinking that we were given the angular velocity \(\omega\) which is in units of \(s^{-1}\) for some reason.
ganeshie8
  • ganeshie8
It makes sense why b, d, f must have same free fall acceleration. Since \(\omega = 0\), the planet rotation wll have no effect on the net force experienced by an object at pole
Kainui
  • Kainui
I was looking at the axis and thinking that axial vector was the angular frequency haha. I was hoping the problem said they had different masses but same kinetic energy, that would have made the problem a little more interesting for those last 3 points.
ganeshie8
  • ganeshie8
|dw:1447480862773:dw|
ganeshie8
  • ganeshie8
|dw:1447480899658:dw|
ganeshie8
  • ganeshie8
Since \(F_{centrifugal}\) and \(F_{gravity}\) are opposing each other, an object at equator feels less net force. did i get it right ?
ganeshie8
  • ganeshie8
the greater the \(\omega\), the greater is the centrifugal force, and lesser the net force felt by the body on the surface of planet at equator
ganeshie8
  • ganeshie8
wish I could get access to the solution manual...
Kainui
  • Kainui
I heard they launch spacecrafts closer to the equator so that they are able to use the centrifugal force to help throw the ship into space by a little extra.
Kainui
  • Kainui
In other words, I don't know anything lol
ganeshie8
  • ganeshie8
it does create the centrifugal force, but this is a pseudo force experienced only by the objects in rotational frame
ganeshie8
  • ganeshie8
suppose you're at the end of a string rotating in circular motion like below : |dw:1447481408663:dw|
ganeshie8
  • ganeshie8
you do feel a force along this direction : |dw:1447481619216:dw|
ganeshie8
  • ganeshie8
you feel that force only if you're in circular motion that centrifugal force disappears the moment you stop rotating
ganeshie8
  • ganeshie8
they tried using that to create artificial gravity in ISS but sooner they discarded the idea due to other health related issues...
Kainui
  • Kainui
Ahhh ok except instead of a string it's gravity, so the faster you're spinning the larger the centrifugal force, which in terms of period T and radius R has acceleration: \[a= \frac{v^2}{r} = \frac{(r/T)^2}{r} = \frac{r}{T^2}\] So shorter period means faster acceleration like we would expect, ok I think this is making more sense to me now.
dan815
  • dan815
"free fall acceleration g at them" ?
ganeshie8
  • ganeshie8
\(a_c = \dfrac{r}{T^2}\) and this is opposing the gravity. so i guess \[g = a_g - a_c\] the greater the centrifugal force\(a_c\), the lesser the free fall acceleration \(g\) hope my understandign isn't flawed...
Astrophysics
  • Astrophysics
He meant where is g the highest and lowest
Astrophysics
  • Astrophysics
Well we have b=d=f
dan815
  • dan815
|dw:1447482182670:dw|
Kainui
  • Kainui
@dan815 Yeah, so in physics you have this equation F=ma and \(F=G\frac{mM}{R^2}\) where G is he gravitational constant, M is the mass of the planet, and R is the radius of the planet. The acceleration \(a\) is given a special name of g when they're equal, \[mg=G\frac{mM}{R^2}\] in other words it's independent of mass of the object being pulled (much like the electric field): \[g=G\frac{M}{R^2}\]
Astrophysics
  • Astrophysics
Omg dan
dan815
  • dan815
what?
dan815
  • dan815
are u guys already past that?
Kainui
  • Kainui
@ganeshie8 Yeah that's what I'm saying too
Astrophysics
  • Astrophysics
Haha, dan that makes a lot of sense because e is rotating the slowest
dan815
  • dan815
well ya the rotating is pushing u away, so when someone jumps he will jump higher on the faster rotating thing
dan815
  • dan815
so that means g is less
Astrophysics
  • Astrophysics
Exactly!
ganeshie8
  • ganeshie8
|dw:1447482306635:dw|