That tower question (1 Viewer)

[AVY]

The mass will go into orbit.

Have a look at this extract from wikipedia:
"Space tower
A space tower is a tower that would reach outer space. To avoid an immediate need for a vehicle launched at orbital velocity to raise its perigee, a tower would have to extend above the edge of space (above the 100 km Kármán line), but far lesser tower height could reduce atmospheric drag losses during ascent. Satellites can orbit temporarily in elliptical orbits dipping as low as 135 km or less, yet orbital decay causing reentry would be rapid unless altitude was later raised to hundreds of kilometers.[20] If the tower went all the way to geosynchronous orbit at approximately 36,000 km, or 22,369 miles, objects released at such height could then drift away with minimal power and would be in a circular orbit. Building a tower to that extreme height is not possible with current materials on Earth. The concept of a structure reaching to geosynchronous orbit was first conceived by Konstantin Tsiolkovsky,[21] who proposed a compression structure, or "Tsiolkovsky tower"."

 

[Stealth_Mel0n]

well i guess the head teacher of physics at my school is wrong ? or maybe multiple answers are accepted along with justification? or maybe the object is dropped instead of thrown? think about a tower not that high (centerpoint tower, assuming its on equator) and assume atmospheric friction has no result, will it orbit in the sky?

You're wrong. Sorry.

 

[someth1ng]

You're wrong. Sorry.

+1

There's not enough tangential velocity at just a tall tower like Sydney Tower. We're talking about a tower as tall as the geostationary orbit.

It's two totally different areas.

 

[Stealth_Mel0n]

+1

There's not enough tangential velocity at just a tall tower like Sydney Tower. We're talking about a tower as tall as the geostationary orbit.

It's two totally different areas.

Yep, I have a feeling that these people just overthought it. It really wasn't that difficult. The one on the equator didn't need to 'go' into geostationary orbit, it already was in geostationary orbit when you think about it, it already had the necessary orbital velocity, orbital period and whatnot. The one on the pole would drop straight down as it does not have the velocity provided by the spinning of the earth as the equator does.

 

[someth1ng]

Yep, I have a feeling that these people just overthought it. It really wasn't that difficult. The one on the equator didn't need to 'go' into geostationary orbit, it already was in geostationary orbit when you think about it, it already had the necessary orbital velocity, orbital period and whatnot. The one on the pole would drop straight down as it does not have the velocity provided by the spinning of the earth as the equator does.

Probably. At first, coriolis effect came to mind and I thought about maintaining the tangential velocity and hence, remains in orbit...

 

[AnimeX]

Hi sorry to rebirth this thread but i'm having difficulty in understanding the whole concept of tower B.

I kind of understand tower A, but say I threw something downwards from tower B, will this mean that it will fall for a bit and just stay stationary after 10m or so?

I a human were to walk out of the tower door (say there was a door), would the human simply be orbiting with the Earth too? so an earth observer below can see him floating above?

Sorry for overthinking this LOL, I tend to do this~

 

[hayabusaboston]

Must include a statement about relative motion to each other to ensure full marks.

 

[iBibah]

Must include a statement about relative motion to each other to ensure full marks.

No you don't. Must include a statement about relative motion to their respective towers but not to each other.