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 indicatedâ€”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.

- ganeshie8

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- ganeshie8

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- Astrophysics

Fc=Fg

- Astrophysics

Well the gravity at b, d, and f would be less than it is at the equator right

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## More answers

- Astrophysics

Sorry would be stronger

- ganeshie8

Since they are ideal spheres, I think the gravity should be same everywhere on the surface ?

- Astrophysics

That's a good point

- 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

there is also centrifugal force on the surface if we switch to rotational frame

- 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

what is coriolis effect ?

- 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

I don't think coriolis effect has anything to do with it

- 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

Well, it means the points at equator are moving faster than points at the pole

- ganeshie8

thats coriolis effect ?

- Astrophysics

Think so man, I barely remember this stuff

- ganeshie8

equator region moving faster than the pole region

- ganeshie8

makes sense... i have just seen a short video on why hurricanes rotate as we speak

- 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

so coriolis effect can be ignored here as there is no movement parallel to longitudes here ?

- Kainui

No I have no idea what the coriolis effect is, so I have no clue

- Astrophysics

I think we can say that, and why'd you say it then kai!

- 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

shouldn't the length of arrows be "decreasing" as a planet with shorter period provides a greater centripetal force ?
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- 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

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

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

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- ganeshie8

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- 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

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

wish I could get access to the solution manual...

- 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

In other words, I don't know anything lol

- ganeshie8

it does create the centrifugal force, but this is a pseudo force experienced only by the objects in rotational frame

- ganeshie8

suppose you're at the end of a string rotating in circular motion like below :
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- ganeshie8

you do feel a force along this direction :
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- ganeshie8

you feel that force only if you're in circular motion
that centrifugal force disappears the moment you stop rotating

- ganeshie8

they tried using that to create artificial gravity in ISS but sooner they discarded the idea due to other health related issues...

- 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

"free fall acceleration g at them" ?

- 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

He meant where is g the highest and lowest

- Astrophysics

Well we have b=d=f

- dan815

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- 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

Omg dan

- dan815

what?

- dan815

are u guys already past that?

- Kainui

@ganeshie8 Yeah that's what I'm saying too

- Astrophysics

Haha, dan that makes a lot of sense because e is rotating the slowest

- dan815

well ya the rotating is pushing u away, so when someone jumps he will jump higher on the faster rotating thing

- dan815

so that means g is less

- Astrophysics

Exactly!

- ganeshie8

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