help with calculation problems (1 Viewer)

[abdooooo!!!]

Originally posted by zeropoint
You wrote that the apparent weight is 2000 N. It's not. Apparent weight is equal to the sum of the reaction forces on a body. In this case 1200 N. We ignore the mg term since it is perpendicular to the direction of motion, and as we well know from projectile motion, the force of gravity has no effect on the horizontal motion.

oh so the apparent wieght in orbit is just the net force and not (mg + ma). did understand it right?

but i think you read the question wrong, because does it not say 'shuttle launch'? or did i not type it correctly

god how do you know all this stuff... you are actually good at the 'real physics'

 

[zeropoint]

Originally posted by abdooooo!!!
oh so the apparent wieght in orbit is just the net force and not (mg + ma). did understand it right?

but i think you read the question wrong, because does it not say 'shuttle launch'? or did i not type it correctly

god how do you know all this stuff... you are actually good at the 'real physics'

Just think of apparent weight as the sum of the reaction forces in a given direction. It makes no sense to talk about ma + mg in this context since the force produced by the engines ma is perpendicular to the force of gravity mg. This is similar to the acceleration of a car. When the car accelerates you experience an apparent weight of ma in the horizontal direction, not ma + mg. This is a strange question however, did you quote the question verbatim? If so, then the answer I gave is correct.

 

[abdooooo!!!]

i think i typed up the exact the question that was given to me... if it is actually as you say then thats just a very strange question. i mean its a trick question isn't it? you just get the same answer twice, why would my teacher want to do that?

 

[wogboy]

The ambiguity here is whether the shuttle is accelerating vertically, or accelerating horizontally. That is:

a) Being lauched from the ground and going up into the sky (i.e. accelerating vertically) or;
b) The shuttle has already been lauched and is up in the sky, and now it needs to be "launched" (fired) horizontally to enter the orbit (i.e. accelerating horizontally).

In the case of a), the answer for the astronaut's apparent weight is 2000 N (mg + ma) because the gravitational force is in the same direction as the acceleration.

On the other hand in the case of b), the answer for the astronaut's apparent weight is 1200N (just ma), because as zeropoint pointed out, the gravitational force has no effect on the weight (since it is perpendicular to the direction of the gravitational force).

I personally believe that a) is the correct case and the answer should be 2000 N (it just seems to make more sense), however I could be wrong.

Are you given any extra information about the launching of the rocket? (e.g. a diagram, or a few sentences) If not then I think the best idea is for you to clear it up with your teacher. Ask your teacher to find out whether the rocket is accelerating vertically from the ground, or accelerating horizontally into orbit.

 

[zeropoint]

Originally posted by wogboy
In the case of a), the answer for the astronaut's apparent weight is 2000 N (mg + ma) because the gravitational force is in the same direction as the acceleration.

Actually, the normal reaction force N = mg is in the same direction as the accceleration, the gravitational force −mg is in the opposite direction.

Originally posted by wogboy

I personally believe that a) is the correct case and the answer should be 2000 N (it just seems to make more sense), however I could be wrong.

I thought this originally, however the question asks for orbital velocity, not launch velocity.

 

[abdooooo!!!]

this is confusing... this is due next tuesday and i have physics on wednesday... oh no

 

[zeropoint]

Originally posted by abdooooo!!!
this is confusing... this is due next tuesday and i have physics on wednesday... oh no

Surely this isn't an assessable assignment? If so, then what are you worried about?

 

[abdooooo!!!]

its worth 10%... but i guess i can include both answers with some explainations about the ambiguity of the question

 

[angmor]

have a question. how would you 'describe where a satellite must be placed in order to be geostationary, using calculations'

they want you to show exact an exact altitude where the satellite must be placed? do you have to use orbital velocity for this?

 

[alcalder]

Geostationary, of course, is stationary over a certain point on Earth, as it rotates. Therefore, the satellite must move fast enough in orbit to:

1. Keep up with the ground below.
2. Not fall back to earth.

Therefore, you need to find an altitude where the centripetal and gravitational forces are equal.

F_grav = Gm₁m₂/r²

F_cent = mv²/r

Therefore:

Gm₁m₂/r² = mv²/r

where m in centripetal equation is mass of satellite = m₂ in grav equation.

Thus,

Gm_earth/r = v²

You know G, you know the mass of the earth. r is the radius of the earth + height of orbit from surface of earth.

Now the satellite must go around the earth at that orbital height once per day - ie the period is T = (1 x 24 x 60 x 60) seconds.

T = 2πr/v
v = 2πr/T

Gm_earth/r = v²

Gm_earth/r = (2πr/T)²

r³ = Gm_earthT²/4π²

Go for it!