# Physics A2-Gravitational FieldsWatch

This discussion is closed.
#1
The orbit of the moon, which has a mass m, is a circle of approx 60R, where R is the radius of the Earth. Mass of Earth is M.

g= GMm/3600R^2

M= 6 x 10^24
R= 6.4 x 10^6

1) Show that the orbital speed of moon around the Earth is approx 1kms^-1.
2) Hence confirm period of orbit.

(I feel as if I should know this but I feel too brain dead at present) Thanks thanks thanks...
0
13 years ago
#2
(Original post by Mimo)
The orbit of the moon, which has a mass m, is a circle of approx 60R, where R is the radius of the Earth. Mass of Earth is M.

g= GMm/3600R^2

M= 6 x 10^24
R= 6.4 x 10^6

1) Show that the orbital speed of moon around the Earth is approx 1kms^-1.
2) Hence confirm period of orbit.

(I feel as if I should know this but I feel too brain dead at present) Thanks thanks thanks...
mV^2/r=GMm/r^2
now V^2=GM/r
r=60R
so v=sqrt(GM/60R)

for 2 the period
T^2=4(pi)^2r^3/GM
or you use V/r=w
w=(2pi/T)
do T=2(pi)r/v
where r=60R and R is the radius of earth
0
#3
(Original post by habosh)
mV^2/r=GMm/r^2
now V^2=GM/r
r=60R
so v=sqrt(GM/60R)

for 2 the period
T^2=4(pi)^2r^3/GM
or you use V/r=w
w=(2pi/T)
do T=2(pi)r/v
where r=60R and R is the radius of earth
Genius.
(I shall called you physicist, as thou wish)
0
13 years ago
#4
(Original post by Mimo)
Genius.
(I shall called you physicist, as thou wish)
yaaaaay hehehehe,though i'm not
0
#5
(Original post by habosh)
yaaaaay hehehehe,though i'm not
You're not?! Well, I shall be the ultimate decider of er... being a physicist...
Answer the ultimate question below and thou shalt be a physicist! *hums* *waves arm dramatically*

Describe how you would show experimentally that the charge stored on a
2.2 x 10^-4 F capacitor is proportional to the potential difference across the capacitor for a range of potential difference between 0 and 15V.

(lol, I can't believe I'm making you answer these questions! But thanks so much anyway!) *hug*
0
13 years ago
#6
remember you decrease the resistance to keep the current constant
0
#7
I don't understand this part:
(Original post by habosh)
fdrow a graph of current time and the charge is the total area under the rectangle graph,also find the vcharge after times that you chosed for the voltage for example if you chosed to reconrd the voltage ever 10 seconds you find the charge after 10 then after 20 ...etc then you drow a graph for the charge against ,the graph would be a stright line through the origin with the gradiant (C) the capacitance
How can you draw a graph of current against time when you are keeping current constant?
0
13 years ago
#8
Its quite sad but I actually read this thread title as gravitational waves - and thought At AS???

Im almost disappointed it wasnt... but in answer to your question its either a straight horizontal line or you arent keeping I constant
0
13 years ago
#9
(Original post by F1 fanatic)
Its quite sad but I actually read this thread title as gravitational waves - and thought At AS???

Im almost disappointed it wasnt... but in answer to your question its either a straight horizontal line or you arent keeping I constant
exactly otherwise it will be an exponantial decaying graph
is my answer to that question right??
0
13 years ago
#10
(Original post by habosh)
exactly otherwise it will be an exponantial decaying graph
is my answer to that question right??
To be honest Im not sure... my memory fails me on the exact details... its not something Ive looked at for a while. I think though that you could just crank up the voltage and monitor the current. That should be a straight line with gradient C as Q=CV

Not totally sure but it rings a bell. I think the experiment you describe is to show the discharge/ charging pattern of a capacitor rather than the charge stored on it. Dont count me on it though
0
13 years ago
#11
Charge up a capacitor to a certain voltage. Disconnect it. Then use a flame probe to measure the electric field around the capacitor. Model said capacitor as an electric dipole, with a pair of point charges seperated by the distance between the plates (that is given these are parallel plate capacitors). Use equation V = 1/(4pi e0 |r|^3) (p . r)
Where p is the seperation vector of the point charges and r is the vector where you measure the voltage
0
13 years ago
#12
Alternatively attach a current source to the capacitor and then plot the voltage over time of the capacitor. You will find that dQ/dt = I. Since I is constant, so is dQ/dt. Hence Q = It and if the graph of V over time is a straight line then Q = cV.
Simple... although I prefer playing with vectors.
0
13 years ago
#13
(Original post by Mehh)
Charge up a capacitor to a certain voltage. Disconnect it. Then use a flame probe to measure the electric field around the capacitor. Model said capacitor as an electric dipole, with a pair of point charges seperated by the distance between the plates (that is given these are parallel plate capacitors). Use equation V = 1/(4pi e0 |r|^3) (p . r)
Where p is the seperation vector of the point charges and r is the vector where you measure the voltage
Electric dipole indeed - show off

This is A level we are talking not degree
0
13 years ago
#14
without compication just record the current and time for several values of V,find Q using It=Q then plot a graph pf Q against V which will be a stright line through the origin with the constant or gradiant C which is the capacitance
0
13 years ago
#15
Yea habosh, you would want to have a way to ensure that It = Q. This relationship only holds given I is a constant. More generally the relationship is I(t)dt = Q [this being an intergration of course]. The point of using a current source is to make sure that I IS constant. Using a voltage source, such as a battery would result in an expoential charging graph as usual (ie the discharge graph upsidedown) and also an expoentially decaying current. Which makes it difficult to measure the charge on the capacitor.
(Original post by F1 fanatic)
Electric dipole indeed - show off

This is A level we are talking not degree
To be completely honest it was the obvious method to measure charge on the plates that way, it was only later I realised you could use a current source and a timer to ensure a known quantity of charge was depositied.
0
#16
Right, ok, after much confusion (with the many ways for doing this), I understand now. Thanks to all. :-)
0
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