Space - DP Problem (1 Viewer)

[Mathematician]

Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler's Law of Periods

where does keplers law come in this ?

this site http://hsc.csu.edu.au/physics/core/space/9_2_2/922net.html#net21
answers it but i dont like that definition and they used kepler's law to come up with V = 2(pi)r/T ? why cant u use v= distance/t where distance = 2(pi) r ?

Can someone else try and do this dot point?

 

[Ragerunner]

Mass of the satellite has no effect on the orbital velocity. Thats about all i know unless i check it up myself which i will do later.

 

[Rahul]

firstly, look at the verb, "define". so you can be assured anything associated wont be a major deal.

It is derived from equating F_g = F_c

V = sqrt(Gm)
......sqrt(r)

The equation shows that the escape velocity depends solely on the Gravitational constant, the mass of the planet and its radius.

then as the csu site shows, you can substitute using keplers third law, although i dont know what that acheives really.

the two main thing you need to know are:
"Orbital velocity is the instantaneous speed (magnitude) in the direction indicated by an arrow (directional) drawn as a tangent to the point of interest on the orbital path."
and the way it is derived. you arent asked to evaluate/analyse, just define.

 

[Mathematician]

I got it now.

Its like this

Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler's Law of Periods

-Orbital velocity is the tangential velocity of a body orbiting a bigger object and undergoing uniform circular motion. Its magnitude never changes, although its direction does.
-It can be derived from Keplers Law of Periods
r^3/T^2 = GM/4pi^2
-But v = 2*(pi)r/T  T= = 2*(pi)r/v
so r^3/(4pi^2r^2/v^2)= GM/4pi^2
r^3v^2/(4pi^2r^2) = rv^2/(4pi^2) = GM/4pi^2
rv^2=GM
v^2= GM/r
v = √ GM/r

Where v is the orbital velocity, G is the gravitational constant equal to
6.67*10^-11 Nm^2kg^-2, M is the mass of the earth and r is the radius of the orbit which is the distance between the two centre of masses which are the satellite and the earth.

- This relationship tells us that if the radius of the satellites orbit is high, we would expect a high orbital velocity and that since the mass of the earth is relatively high compared to the satellite, the orbital velocity would be quite significant.
- It also tells us that the orbital velocity depends only on the universal gravitational constant, Mass of the earth and the orbits radius.


Extra (derivation not relevant to dot point)
-Alternatively it can be derived by the centripetal force acting on the satellite (in this case) which is supplied by the gravitational attraction between the earth and the satellite.
mv^2/r = GMm/r^2
v^2 = GM/r
v = √ GM/r

 

[Comedy_Al]

Sounds right to me

 

[Ragerunner]

looking at those formulas mathematician wrote scares the living hell out of me. Thank god physics is more english based now

 

[Dash]

I hate trying to make sense of all these calculations that are posted up here on BOS...
They give me a headache...

Can't BOS make a equation maker thingy for maths based courses???
It would be heaps helpful...

 

[Rahul]

just a tip, when you are trying to help/answer a question. the best way is to write in on a piece of paper and then write it up on the comp. or when you are trying to make sense of it all copy it down on a piece of paper and then try to work with it. it get tedious at times, no doubt about it.

or people can just use the maths equation thingo in word and attatch the file, or if it is small enough a pic, so u dont need to open it.

 

[NSBHSchoolie]

Re: I got it now.

Originally posted by Mathematician
- This relationship tells us that if the radius of the satellites orbit is high, we would expect a high orbital velocity and that since the mass of the earth is relatively high compared to the satellite, the orbital velocity would be quite significant.

Sorry I have a problem. I don't think that orbital velocity does not increase with radius as you asserted.

Since v is inversely proportional to r, it would follow that as r increases, v decreases. As r->infinity, v->0. This makes sense if you think about it. Gravity gets weaker with distance.

 

[Mathematician]

...

oh shit. Sorry my mistake. You are right, and i meant to write down what u said, but got tired .... lol

btw u make errors too lol
I don't think that orbital velocity does not increase with radius as you asserted

u meant
I don't think that orbital velocity does increase with radius as you asserted.

I THINK??



Thanx anyway.

Some of my notes was going to be WRONG!!! LOL

And i got another q.

With this dotpoint
Discuss issues associated with safe re-entry for a manned spacecraft into the Earths atmosphere and landing on the Earths surface.

Can we just talk about the Spaceshuttle???

Whats the difference between the tiles used in the spaceshuttle and the Apollo space capsule or other spacecraft?

 

[Ragerunner]

Correct me if im wrong but i think the early spacecrafts used ablative materials that vaporise upon re-entry and absorb the heat.

The modern space shuttle uses ceramic/fiberglass tiles that are attached to the ship and is re-usable

 

[Rahul]

Discuss issues associated with safe re-entry for a manned spacecraft into the Earths atmosphere and landing on the Earths surface.

*the angle of reentry has to be between 5.2(degrees) and 7.2(degrees). too shallow and the ship will bounce off into the atmosphere and if it is too deep then it will burn up in the atmosphere due to friction

*the spacecraft zig-zags in the atmosphere to reduce its speed and therefore friction

*ablative material is used on the underside of the cratft. this material vapourisis, so the heat doesnt get conducted to the metal beneath it.

*a blunt shaped craft is made. this way a shockwave barrier is made while it travels at great speeds through the atmosphere, reducing the heat generated due to friction. this barrier absorbs the heat.

*astronauts lie on their back so that the g-forces are directed to their backs. this reduces the chances of black-outs(when blood rushes to their feet due to acceleration), red-outs(when blood rushes to their eyes due to acceleration) or even fatalities.

*communication is lost due to an ionisation blackout. spacecraft gets surrounded by ions due to the heat generated by the friction.

hmmm...thats all i can think of, there will probably be more.

Whats the difference between the tiles used in the spaceshuttle and the Apollo space capsule or other spacecraft?
i am not too sure, it isnt part of the syllabus, i dont think

edit/ i think ragerunner has answered that question well.

 

[Mathematician]

...

I think u are correct.

Can u plz explain to me why they are re-usable?

And in the textbook it says the older spacecrafts used 'ablative' material , made of fibreglass also????

 

[jims]

on the apollo missons, the ablative material was an aluminium honeycomb into which phenolic epoxy resin has been injected and cured. most other missions used much the same sorta heat shield which could only be used once.

the spaceshuttle, however, has those ceramic/fibreglass tiles which are generally not damaged on reentry because they can withstand the heat without being vapourised, charred or weakened.