MedVision ad

G - forces - Help Needed (1 Viewer)

eX-Bhai

Member
Joined
Mar 15, 2005
Messages
78
Gender
Male
HSC
1998
I kinda understand the concept of g-forces... but its just the formula which is putting me off...

I understand, that g-forces = apparent weight / true weight

apparent weight = mg + ma
true weight = mg

therefore g-forces = (g + a) / g

where g = acceleration due to gravity and
a = acceleration of the object

by looking at this formula, you can see that the g-force is only dependent on the acceleration due to gravity and the acceleration of the object. Therefore making mass, time and displacement is independent right?..

But I looked at one of the eddasist questions:
"Explain why the g-forces increase as the altitude of a rocket increases"

But shouldn't the g-force remain constant as displacement, or altitude, is independent?

One of my friends came with this answer:

increase in altitude --> decrease in air particles (therefore less friction) ---> increase in acceleration --> increase in g-force.

also

a rocket loses mass due to the consumption of fuel --> increased acceleration --> increased g-force

well I believe that her explanation is right.. but now im kinda caught in the middle saying that mass and displacement is independent to the g-force..

so yeh, if anyone could clarify that mass and displacement is independent with an explanation of my friends answer - it would be greatly appreciated. Thank you in advanced :)

~eX-Bhai
 

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
eX-Bhai said:
a rocket loses mass due to the consumption of fuel --> increased acceleration --> increased g-force
That statement is valid. As the rocket continues in it's path away from the earth, the fuel is rapidly consumed, the mass of the rocket decreases quickly and the rocket gradually continues to accelerate. Therefore there is an increased g-force on the occupants of the rocket. In effect, you could consider that the change in displacement from the earth does play a role in affecting the g-force; as you get further away, you are consuming more fuel, making the rocket lighter as you get further away and increasing the acceleration and g-force.
 

eX-Bhai

Member
Joined
Mar 15, 2005
Messages
78
Gender
Male
HSC
1998
Riviet said:
That statement is valid. As the rocket continues in it's path away from the earth, the fuel is rapidly consumed, the mass of the rocket decreases quickly and the rocket gradually continues to accelerate. Therefore there is an increased g-force on the occupants of the rocket. In effect, you could consider that the change in displacement from the earth does play a role in affecting the g-force; as you get further away, you are consuming more fuel, making the rocket lighter as you get further away and increasing the acceleration and g-force.
Yes, I understood that concept, but i was wanting to know how that relates to the formula... the formula clearly reveals that mass, displacement and time is independent. The only dependent variables are the acceleration due to gravity and the acceleration of the object. That is where I am confused - relating the formula to this statement.
 

zeropoint

Member
Joined
Dec 10, 2003
Messages
243
Location
Sydney
Gender
Male
HSC
2004
eX-Bhai said:
Yes, I understood that concept, but i was wanting to know how that relates to the formula... the formula clearly reveals that mass, displacement and time is independent. The only dependent variables are the acceleration due to gravity and the acceleration of the object. That is where I am confused - relating the formula to this statement.
Different masses.

G-force is independent of _your_ mass, not the mass of the rocket. Note that G-force is a function of a, which is in turn a function of the mass of the rocket, making G-force dependent on the mass of the rocket.
 

zeropoint

Member
Joined
Dec 10, 2003
Messages
243
Location
Sydney
Gender
Male
HSC
2004
Another thing. The gravitational acceleration g is a function of altitude h via:

g = GM_earth /(r_earth + h)^2

therefore g-force is also a function of altitude.
 

eX-Bhai

Member
Joined
Mar 15, 2005
Messages
78
Gender
Male
HSC
1998
zeropoint said:
Another thing. The gravitational acceleration g is a function of altitude h via:

g = GM_earth /(r_earth + h)^2

therefore g-force is also a function of altitude.
Where did you get that formula from?
 

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
eX-Bhai said:
Where did you get that formula from?
It's the formula to find g on any planet at a particular altitude from the planet.

If you are on the surface of the planet, the equation is simply g=GM/r2, where M is the mass of the planet and r is the radius of the planet. Therefore when you are a certain distance away from the surface of the planet and you want to find gravity at that point, you need to add the altitude (h) to the radius of the planet when applying the formula. The equation then becomes: g=GM/(r+h)2

Does that make sense? :)
 

Ioup

Member
Joined
Oct 24, 2005
Messages
73
Gender
Male
HSC
2006
An easy way to understand this is to use the equation
a=T-mg/m
Since a rockets thrust remains essentially the same in flight but its mass decreases the acceleration will be greater towards the end of its flight.
Like Zeropoint said the mass of the rocket does affect the G force.
 

zeropoint

Member
Joined
Dec 10, 2003
Messages
243
Location
Sydney
Gender
Male
HSC
2004
eX-Bhai said:
Where did you get that formula from?
The formula is a consequence of Newton's law of universal gravitation, which says that a test paticle m experiences a force F from a body of mass M given by

F = G M m / r2

where r is the distance between the centres of mass. If the test particle is subject to no other forces than gravity (in free-fall), then applying Newton's second law to the test mass we obtain

m a = G M m / r2

and thus the gravitational acceleration is

g = a = G M / r2

for an object at an altitude h above the surface of the earth, the distance between the centre of mass of the earth and the object is r = h + Rearth.
 

eX-Bhai

Member
Joined
Mar 15, 2005
Messages
78
Gender
Male
HSC
1998
Hm.. I'm not sure if ive grasped the concept, but can somebody please clarify my understanding of g-forces through this attachment? Thanks in advice.

Well I haven't really answered part two of the question, but according to the surfing physics sheet on g-forces, "it is often easier to just calculate the acceleration of the rocket in terms of multiples of 9.8 as this effectively gives the same answer."


"Remember, we still need to add 1 to our answer here - assuming the rocket is still in the Earth's gravitational field" - Whats the use in adding one to the g-force?

On top of this, I'm kind of confused with their example.

What it says is that:

-----------------------------------------

Example: A 500kg rocket carrying 300kg of fuel develops a thrust of 10 000N. Calculate the maximum g force that would be experienced by its astronaut.

Data:
m = 500 kg
thrust = 10 000 N
fuel = 300 kg

Calculation:
Maximum acceleration experienced just as fuel is exhausted, so:

a = (10000 - 200 x 9.8)/200
= 40.2

so, gforce due to acceleration = 40.2/9.8
= 4.1
therefore, total gforce experienced = 5.1g

------------------------------------------------------------

Where did he get that formula from? a = (10000 - 200 x 9.8)/200? I'm just really confused =x
 
Last edited:

Ioup

Member
Joined
Oct 24, 2005
Messages
73
Gender
Male
HSC
2006
eX-Bhai said:
-----------------------------------------

Example: A 500kg rocket carrying 300kg of fuel develops a thrust of 10 000N. Calculate the maximum g force that would be experienced by its astronaut.

Data:
m = 500 kg
thrust = 10 000 N
fuel = 300 kg

Calculation:
Maximum acceleration experienced just as fuel is exhausted, so:

a = (10000 - 200 x 9.8)/200
= 40.2

so, gforce due to acceleration = 40.2/9.8
= 4.1
therefore, total gforce experienced = 5.1g

------------------------------------------------------------

Where did he get that formula from? a = (10000 - 200 x 9.8)/200? I'm just really confused =x
I will mention again that you can use the equation
a=T-mg/m
Where T is the thrust in Newtons...
Since G force is measured by multiples of 9.8.
divide the acceleration of the rocket given by the a=t-mg/m by 9.8 to get the G force.
I have the sheets from that textbook given by the teacher but not the answers.
Anyone know where I can buy the textbook surfing physics from?
 

eX-Bhai

Member
Joined
Mar 15, 2005
Messages
78
Gender
Male
HSC
1998
Ioup said:
I will mention again that you can use the equation
a=T-mg/m
Where T is the thrust in Newtons...
Since G force is measured by multiples of 9.8.
divide the acceleration of the rocket given by the a=t-mg/m by 9.8 to get the G force.
I have the sheets from that textbook given by the teacher but not the answers.
Anyone know where I can buy the textbook surfing physics from?
Ohhhh I remember this formula now.. yeh, my teacher said that its a good one to further your understanding - but its not required in the syllabus
 

Ioup

Member
Joined
Oct 24, 2005
Messages
73
Gender
Male
HSC
2006
As for the addition of an extra G force I'm a little bit confused on that.
I use a different textbook to "Surfing" and the G force calculations dont require an extra 1 g...
 

Mountain.Dew

Magician, and Lawyer.
Joined
Nov 6, 2005
Messages
825
Location
Sydney, Australia
Gender
Male
HSC
2005
considering this:

-----------------------------------------

Example: A 500kg rocket carrying 300kg of fuel develops a thrust of 10 000N. Calculate the maximum g force that would be experienced by its astronaut.

Data:
m = 500 kg
thrust = 10 000 N
fuel = 300 kg

Calculation:
Maximum acceleration experienced just as fuel is exhausted, so:

a = (10000 - 200 x 9.8)/200
= 40.2

so, gforce due to acceleration = 40.2/9.8
= 4.1
therefore, total gforce experienced = 5.1g

------------------------------------------------------------
realise in the line "
so, gforce due to acceleration = 40.2/9.8
= 4.1", this was according to the formula g-force = a/9.8 --> this did not take into consideration the "g" in the standard formula --> g-force = (a + g )/ 9.8

BUT we do realise that we add 1 because g/9.8 = 9.8/9.8 = 1. thats why we added one.

IF we considered g-force = (a + g )/ 9.8 and did this --> gforce due to acceleration = 40.2 + 9.8/9.8
= 4.1 + 1 = 5.1, then theres no need for the extra 1g.

my advice --> just follow the general g+a / 9.8 formula --> it works

NOTE: g = 0 in orbit and/or beyond orbit.
 

mitsui

мιтэuι
Joined
Aug 31, 2005
Messages
1,191
Location
somewhere
Gender
Female
HSC
2006
while on the subject,

i understand the concept of g-force

but the dot pt asks "why is the term of g-force used"??

can someone help me on that?

is it becoz it is easier to caculate becoz wifout g-force, the force measurement is way too big? o_O
 

frangipane

New Member
Joined
Nov 26, 2005
Messages
13
Location
Sydney...Were else???
Gender
Female
HSC
2006
Im stuck on the same point too...
Have you gone onto HSC ONLINE-Physics? I went and it gave a reasonable explaination.

It says that its used because true weight cannot be felt yet apparent weight is what the astronaut experiences. 'G-force' is the expression of the proportion of one to the other...if that makes sense. Im still looking for a better explaination.

If you find one, PLEASE POST!!
 

mitsui

мιтэuι
Joined
Aug 31, 2005
Messages
1,191
Location
somewhere
Gender
Female
HSC
2006
well.
wif the hint of ex_bhai

and help of the excel physics hsc book. this sounds a bit more reasonable:

human bodies can undergo large constant speed or velocity wifout feeling any stress (ie. when u r on a plane etc.)
but when u r undergo a changing speed (ie. undergo acceleration), ur body is under "acceleration stress"

g-force is a measurement to purely measure the acceleration an astronut is under to better help understand the amount of stress they would b under (to help take precautions??)

anyhow. the max. limite that humans go b under is 4 or 5 g-forces (4 says macquarie, while 5 says excel)..--"

hope this will muffle its way thru the question. >_>
 

eX-Bhai

Member
Joined
Mar 15, 2005
Messages
78
Gender
Male
HSC
1998
okay, ive got the answer :D ...well i THINK i got the answer.. anyway =P

First of all, we know that in order to achieve an escape velocity.. large accelerations need to be achieved right?? These accelerations create forces upon an astronaut.. which are known as g-forces. And we know that g-forces are closely related to acceleration (increase in g-force --> increase in acceleration). So anyway, according to the excel book, humans can withstand large velocities/speeds, but they cannot stand large changes in speed (ie accelerations), as it causes 'acceleration stress' to the human body. This is why g-forces are used to explain the forces acting on the astronaut during a launch.

If you're still kinda confused, try to consider the analagy that ur in a rollercoaster. As the rollercoaster initially launches, you feel a big queezy right? Thats because of its accelerations.. But once it hits a relatively constant velocity and zero acceleration, you feel at ease.. thats because there is no acceleration..

I hope this helps..

~eX-Bhai
 

Mountain.Dew

Magician, and Lawyer.
Joined
Nov 6, 2005
Messages
825
Location
Sydney, Australia
Gender
Male
HSC
2005
mitsui said:
while on the subject,

i understand the concept of g-force

but the dot pt asks "why is the term of g-force used"??

can someone help me on that?

is it becoz it is easier to caculate becoz wifout g-force, the force measurement is way too big? o_O
mmmm u could say its just for convenience, and we as humans can relate to ****gs because we know that what we are experiencing now is 1g. <-- its all a relative thing, to put the reality of space travel and rocket launches in our perspective.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top