# Circular motion/ orbital speed help

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#1
2ai).

So I originally tried using v^2 = GM/r

which the markscheme give a mark for however, I'm not sure of the mass?

and the other way is g = v^2 / r

If someone can explain the first way and also where the second equation is from?

Thanks

http://www.ocr.org.uk/Images/144792-...nian-world.pdf
0
3 years ago
#2
Look at the units.

G is ms^-2
V^2 is m^2s^-2
r = m

The mass cancels 1 time gives you ms^-2

Radius is from centre of both. so 6400+800=7200km
You should memorise the mass of the sun and earth. Earth is 6x10^24kg so you can always use it as an approach if you struggle.
square root radius x field strength
0
3 years ago
#3
and for 2aii

v/r=w
T=2pi/w

T=2pi.r/w

Input values and you get your time period
0
3 years ago
#4
Whenever you do physics. If you struggle with a question which will happen to everyone always go back to the units. Every formula is homogeneous and so you can work around units with what they give you. Try to learn to derive your own formula, hence why the mark scheme tend to show more than 1 method as well as more equations that are not generally shown on your data sheet. This will help a lot when you get given questions that combine multiple topics and they can throw you off when they give you what you may think useless information. Always look at the units put them in base units and start cancelling things out to try get to the unit you need. This can be a life saver in exams where if you fail to catch on to the question you can still get the answer without really knowing how to do it. Just a tip.
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#5
(Original post by Konanabanana)
Look at the units.

G is ms^-2
V^2 is m^2s^-2
r = m

The mass cancels 1 time gives you ms^-2

Radius is from centre of both. so 6400+800=7200km
You should memorise the mass of the sun and earth. Earth is 6x10^24kg so you can always use it as an approach if you struggle.
square root radius x field strength
Nah you have lost me. Forget about all this unit stuff as I've never seen it like that.

From the way they have it in the mark scheme

v^2/r = GM

yet I still don't understand where

g= v^2/r is from
0
3 years ago
#6
(Original post by Super199)
Nah you have lost me. Forget about all this unit stuff as I've never seen it like that.

From the way they have it in the mark scheme

v^2/r = GM

yet I still don't understand where

g= v^2/r is from

v^2/r = GM

X both sides by r to give v^2= Gmr

yet I still don't understand where

g= v^2/r is from

Like i said there are more than 1 approach to answering questions but ultimately they all reach the same answer as the units all cancel in a way. If you want to actually understand formula in physics u really need to learn your units and the derivatives of them in base units.

If you have any other questions in physics of any topic i'm more than happy to help.
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#7
(Original post by Konanabanana)
v^2/r = GM

X both sides by r to give v^2= Gmr

yet I still don't understand where

g= v^2/r is from

Like i said there are more than 1 approach to answering questions but ultimately they all reach the same answer as the units all cancel in a way. If you want to actually understand formula in physics u really need to learn your units and the derivatives of them in base units.

If you have any other questions in physics of any topic i'm more than happy to help.
Ok thanks but try have a go and plus in your values for G,M and r and tell me if you get 7400 as an answer
0
3 years ago
#8
(Original post by Super199)
Ok thanks but try have a go and plus in your values for G,M and r and tell me if you get 7400 as an answer
Yeah i got 7455.42

Squar root that number and it gives 7455.42m/s

Do you not know what each symbol means?
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3 years ago
#9
I told you if you memorise the mass of the sun and earth you can use gravitational constant to help work out your answers an alternative way. It's really crucial
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3 years ago
#10
sun may change between exam board btu i learn it as 2x10^30
Earth may change slightly between exam boards i learnt it as 6x10^24
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#11
(Original post by Konanabanana)
Yeah i got 7455.42

Squar root that number and it gives 7455.42m/s

Do you not know what each symbol means?
I thought it was v^2 = GMr

v^2 = (6.67 * 10^-11) x (7200 x 10^6) x ( 6x10^24)

Square rooting that does not give me 7455

oh ffs I got it now. It was Gm/r

Another question 3bii)
0
3 years ago
#12
(Original post by Super199)
I thought it was v^2 = GMr

v^2 = (6.67 * 10^-11) x (7200 x 10^6) x ( 6x10^24)

Square rooting that does not give me 7455

oh ffs I got it now. It was Gm/r

Another question 3bii)
Been a year since i've done physics, give me like few mins to work through the previous questions.
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3 years ago
#13
Ah 3bii is so hard to explain. It'svery similar to how in cosmology topics like intensity and flux. For example the flux on mars compared to earth. You have to show a relationship using inverse square law. This question is a similar approach.

You need to know which is going to be larger the gravitation field strength for M or A. It's larger for A than M. And do the same for radius^2
Radius for A is smaller than that of M. You know g proportional to r^2
Mg/Ag=r^2A/r^2M

This is inverse square law that's why its a fraction. So if you increase g then you decrease r. So you know if g for M is smaller then you know r is bigger.

Really hard to explain it in messages.

Try watching a video on youtube by drAphysics and watch his videos on planterary science
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#14
(Original post by Konanabanana)
Ah 3bii is so hard to explain. It'svery similar to how in cosmology topics like intensity and flux. For example the flux on mars compared to earth. You have to show a relationship using inverse square law. This question is a similar approach.

You need to know which is going to be larger the gravitation field strength for M or A. It's larger for A than M. And do the same for radius^2
Radius for A is smaller than that of M. You know g proportional to r^2
Mg/Ag=r^2A/r^2M

This is inverse square law that's why its a fraction. So if you increase g then you decrease r. So you know if g for M is smaller then you know r is bigger.

Really hard to explain it in messages.

Try watching a video on youtube by drAphysics and watch his videos on planterary science
http://www.ocr.org.uk/Images/142403-...ld-january.pdf

Do you get what the mark scheme has done?
0
3 years ago
#15
(Original post by Super199)
http://www.ocr.org.uk/Images/142403-...ld-january.pdf

Do you get what the mark scheme has done?
Yh there relationship is the same as mine. then they just input the values they were given as well as the value worked out prior. Seems pretty good to me. This kind of question used to always trick people. Most people in my class when were learnt it either always got it right or either always got it wrong. The good thing though is its always the same answer for these kind of questions the relationship is the part which people get wrong. So if you know that part the in putting numbers is easy of course. anything else?
0
3 years ago
#16
For 2ai)
g=F/m
Therefore: F=gm
F=(mv^2)/r
Equating these gives gm=(mv^2)/r
So v^2=gr as the two m's cancel. This gives v=7445ms^-1, which is 7400ms^-1 to 2sigfig (unlike the other guy's answer)

For 3bii) g=-GM/r^2
-GM is constant, so g is proportional to 1/r^2
Therefore g[at M] / g[at A]=(1/r[at M]^2) / (1/r[at A]^2)
So to find gravitational field strength at M, g[at M]:
g[at M] = g[at A] * ((1/(orbital radius at M)^2) / (1/(orbital radius at A)^2))
= 7.5 * ((1/(2.4*10^10)^2)/(1/(1.3*10^8)^2))
= 2.22*10^-4 Nkg^-1
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#17
(Original post by john475869_)
For 2ai)
g=F/m
Therefore: F=gm
F=(mv^2)/r
Equating these gives gm=(mv^2)/r
So v^2=gr as the two m's cancel. This gives v=7445ms^-1, which is 7400ms^-1 to 2sigfig (unlike the other guy's answer)

For 3bii) g=-GM/r^2
-GM is constant, so g is proportional to 1/r^2
Therefore g[at M] / g[at A]=(1/r[at M]^2) / (1/r[at A]^2)
So to find gravitational field strength at M, g[at M]:
g[at M] = g[at A] * ((1/(orbital radius at M)^2) / (1/(orbital radius at A)^2))
= 7.5 * ((1/(2.4*10^10)^2)/(1/(1.3*10^8)^2))
= 2.22*10^-4 Nkg^-1
Bit in bold?
How did you come to that relationship
Why wouldn't it be like Gfield A/ r^2A = Gfield M / r^2m
0
3 years ago
#18
The ratios are the same

g is proportional to 1/r^2, so can be written as:
g=k/r^2 where k is a constant

So for points M and A
g[M]=k/r^2[M] and g[A]=k/r^2[A]

Rearrange both equations for k, and let them equal each other
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