# Circular motion/ orbital speed help

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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

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

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#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

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

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#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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(Original post by

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

**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

From the way they have it in the mark scheme

v^2/r = GM

Solve for v please.

yet I still don't understand where

g= v^2/r is from

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#6

(Original post by

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

Solve for v please.

yet I still don't understand where

g= v^2/r is from

**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

Solve for v please.

yet I still don't understand where

g= v^2/r is from

v^2/r = GM

Solve for v please

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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(Original post by

v^2/r = GM

Solve for v please

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.

**Konanabanana**)v^2/r = GM

Solve for v please

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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#8

(Original post by

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

**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

6.67x10^-11x6x10^24/7200000

Squar root that number and it gives 7455.42m/s

Do you not know what each symbol means?

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#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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#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

Earth may change slightly between exam boards i learnt it as 6x10^24

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(Original post by

Yeah i got 7455.42

6.67x10^-11x6x10^24/7200000

Squar root that number and it gives 7455.42m/s

Do you not know what each symbol means?

**Konanabanana**)Yeah i got 7455.42

6.67x10^-11x6x10^24/7200000

Squar root that number and it gives 7455.42m/s

Do you not know what each symbol means?

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

My bad

Another question 3bii)

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#12

(Original post by

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

My bad

Another question 3bii)

**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

My bad

Another question 3bii)

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#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

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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(Original post by

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

**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

Do you get what the mark scheme has done?

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#15

(Original post by

http://www.ocr.org.uk/Images/142403-...ld-january.pdf

Do you get what the mark scheme has done?

**Super199**)http://www.ocr.org.uk/Images/142403-...ld-january.pdf

Do you get what the mark scheme has done?

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#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

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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(Original post by

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

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

**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

How did you come to that relationship

Why wouldn't it be like Gfield A/ r^2A = Gfield M / r^2m

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#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

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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