Calculating Artificial Gravity for Different Radii on a Space Station

[blumfeld0]

a space station is composed of 2 concentric circles. the inner circe has diameter 5km(r=2.5 km) and the outer has a diameter 10 km (r=5km). g= 1.64 m/s^2 for the outer circle and i need g for the inner circle

i know g = G *M/r^2
so perhaps i can calculate M?
1.64 = 6.67E-11 * M/5000^2

so M= 6.1E17 kg for the whole planet.

but how do i calcaulte g only for the inner radius since the mass contained in the inner radius is unknown.
thanks

 

[cesiumfrog]

Wow. You successfully found an equation that contains some of the variables you're interested in. That alone isn't physics.

You calculated the mass of a 10km diameter planet with a particular surface-gravity value.

Then you calculated what that planet's surface-gravity would be if it collapsed to half its previous diameter.

But is mass what causes the apparent gravity on your space-station?

 

[blumfeld0]

obviously its wrong but its all i got. so i calculated the mass of the whole planet. but obviously the mass contained within the inner circle is different. so what do i do?

 

[tim_lou]

I am confused by your question. Is it a space station, or a planet? is the space station in space? on a planet? is it rotating? are you calculating gravity caused by a planet? or by the space station?

 

[blumfeld0]

hello. well in the question there is nothing at all about rotation which is confusing because i thought we needed rotation for artificial gravity. we do not know the mass of the space station.
all i have is a space station made of two concentric circles. i have radius of the inner circle (2500 meters) and the radius of the outer circle (5000 meters)
lastly i know the gravitational acceleration of the outer circle (1.64 m/s^2) which is equivalent to the moons gravitation acceleration.
i need the graviational acceleration (artifcial gravity) of the inner circle.
there is absolutely no other information given in the question.
but yeah i assume the space station is orbiting something in space.

thank you

 

[blumfeld0]

anybody? any ideas?

 

[tim_lou]

If I clearly understand the problem, maybe I can help.

So, is there a diagram associated with the problem? what is the space station's position in relation with the object that is causing the gravitational pull? can I assume that the gravity is constant alone the circles?

if the space station wraps around a planet symmetrically, then you can simply apply the ratio of inverse r^2.

or, if the space station is rotating, then you can apply the centrifugal force in non-inertial frame.
the equation is:F=omega^2*r
(assuming that omega vector is orthogonal to the r vector)

 

[Saketh]

how do i calcaulte g only for the inner radius since the mass contained in the inner radius is unknown.

You don't need the mass. Think about centripetal acceleration.

I think that the question has a space station spinning by itself in space. The outside ring has a different centripetal acceleration than the inside ring.

 

[ShawnD]

You don't need the mass. Think about centripetal acceleration.

I think that the question has a space station spinning by itself in space. The outside ring has a different centripetal acceleration than the inside ring.

The interesting part of this question is that it doesn't even need to tell you if this "gravity" is caused by gravity or by it being a giant centrifuge.
The answer is probably 0.41m/s^2

The entire question is depending on radius because everything else stays constant.

or, if the space station is rotating, then you can apply the centrifugal force in non-inertial frame.
the equation is:F=omega^2*r

Replace the "F" with "a". This raises a good point though, you need to use the formula as omega^2 and r, not as v^2 and r. V changes depending on radius, while omega is not related to radius.

 

[blumfeld0]

i think i got it
a=g = v^2/r = .82 m/s^2
because v^2 = 1.64 * 2500 = g r
r= 5000 m

i don't really understand why the r's are different? any ideas?thanks

 

[OlderDan]

i think i got it
a=g = v^2/r = .82 m/s^2
because v^2 = 1.64 * 2500 = g r
r= 5000 m

i don't really understand why the r's are different? any ideas?


thanks

The radii are different becaue the author of the problem wants you to demonstrate your understanding of how centripetal force depends on radius at a constant angular velocity. (You must make the reasonable assumption that the whole station rotates at one angular velocity to do the problem.) The magnitudes of the g values in the problem are the centripetal accelerations at the two different radii; one is given to establish the angular velocity of the station, which is then used to find the accleration at the second radius. You need not actually compute the angular velocity if you set up a proportion based on the relationship between angular velocity, radius, and centripetal acceleration.