That tower question (4 Viewers)

[Dash8]

I said that the object dropped from the tower on the north pole will fall straight down cause the earth spins on its axis at that point and the object world have no orbital velocity to keep it in orbit. For the other tower, from where the object is dropped, it is already orbiting the earth at the same speed the earth rotates so therefore it will remain in a geostationary orbit.

Pretty retarded question though....

 

[GoldyOrNugget]

Both where at Geostationary points. If one stayed up, so would the other one.

Don't think so. Mass A one was at a geostationary height, but it was on the vertical axis -- i.e. it wasn't orbiting, only rotating. It would thus plummet to the earth. However, Mass B was rotating with the tower, i.e. rotating with a period of 24h. So if the tower were to disappear, it would continue orbiting in geostat orbit.

 

[LifeBoats]

Don't think so. Mass A one was at a geostationary height, but it was on the vertical axis -- i.e. it wasn't orbiting, only rotating. It would thus plummet to the earth. However, Mass B was rotating with the tower, i.e. rotating with a period of 24h. So if the tower were to disappear, it would continue orbiting in geostat orbit.

But that would assume that the initial velocity of Mass B is enough to sustain a geostationary orbit, right?

 

[barbernator]

what goldy is saying is perfectly correct.

 

[Automatia]

v=d/t

geostationary velocity = geostationary distance (given) / geostationary period (from earths 24 hour rotation period)

 

[lance687876]

you have to think about the general concept of satellites.
the satellites in a geostationary orbit have nothing propelling them. therefore mass B will stay in a geostationary orbit. it has the same velocity as the roation of the earth and therefore the same period
l think you also had to talk about the centripetal acceleration B has towards the earth, keeping it there
it's a 4 marker after all.

 

[RishBonjour]

best question I've seen in the HSC. hsc physics needs more shit like this rather than "impact of transistors"

 

[IceDingo]

I reckon Goldy/Automatia/lance/sylock(heythere) are correct. The mass from tower B would stay in orbit as it is released at geostationary orbit height at geostationary orbit velocity. And the one from A would just drop as it has no orbital velocity.

hsc physics needs more shit like this rather than "impact of transistors"

I agree.

 

[egnaro315]

wait, didn't it mention that were at the same altitude not orbital radius or am I wrong

 

[kiinto]

I reckon Goldy/Automatia/lance/sylock(heythere) are correct. The mass from tower B would stay in orbit as it is released at geostationary orbit height at geostationary orbit velocity. And the one from A would just drop as it has no orbital velocity.

That doesn't make sense. The earth's rotational velocity has nothing to do with a satellites orbital velocity (V = sqrt(GM/r).

Orbital velocity would be produced by the inward acceleration due to gravity, as gravity is a centripetal force.

Both satellites have the same orbital period (of 24 hours) unless of coarse you said that equatorial bulge made height of towers different --> Orbital period less at North Pole. However, only satellite at earth's equator will appear to be motionless in the sky.

 

[LlamaBoi]

We should still get 2 marks if we said the mass from Tower A dropped straight down, right?

 

[lance687876]

We should still get 2 marks if we said the mass from Tower A dropped straight down, right?

2 marks? for that? a 10 year old kid knows that. what do you think the hsc is? lol

 

[Sindivyn]

We should still get 2 marks if we said the mass from Tower A dropped straight down, right?

If you mentioned that it would accelerate at an accelerating rate, probably. (The value of g increases as the object falls).

 

[Automatia]

That doesn't make sense. The earth's rotational velocity has nothing to do with a satellites orbital velocity (V = sqrt(GM/r).

Orbital velocity would be produced by the inward acceleration due to gravity, as gravity is a centripetal force.

Both satellites have the same orbital period (of 24 hours) unless of coarse you said that equatorial bulge made height of towers different --> Orbital period less at North Pole. However, only satellite at earth's equator will appear to be motionless in the sky.

No. You need to think about the way the Earth rotates. Its spins on an axis roughly through the poles. Therefore, by hsc physics, the ball on the tower at the poles has no velocity and merely falls down to Earth. However, the ball on the tower at the equator has been rotating around with the earth at the same radius and velocity as a geostationary orbit and so when it is released it stays in a geostationary orbit.

 

[Fizzy_Cyst]

Awesome question.

Definitely will discriminate the band 6 students from the band 5 students.

 

[RishBonjour]

Awesome question.

Definitely will discriminate the band 6 students from the band 5 students.

what do you think the answer was?

 

[gayforniall]

For Tower A I said it was a Polar Orbit... and for Tower B I said it functions as a Geostationary Orbit.

 

[lance687876]

For Tower A I said it was a Polar Orbit... and for Tower B I said it functions as a Geostationary Orbit.

LOOOOL polar orbit?
dude, that's for asynchronous satellites close to the surface, travelling at insane speeds.

 

[learninggg]

i talked about how earth isn't a perfect sphere, so the gravitational acceleration is different around the earth. tower A would of reached the surface first .. i think.

+1

 

[someth1ng]

Pretty sure it was

Tower A: It fell right onto the north pole

Tower B: It stayed at the point of release, acting as a geostationary satellite would.

Explain a bit on why Tower A is not geostationary... should be 4. Too bad I'll be lucky to get one :O

This is what I did and is definitely correct.