AQA A2 Physics Gravitational Fields help

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kirino1
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Im stuck on this question, any help would be greatly appreciated

The moon has a radius of 1740km and its surface gravitational field strength is 1.62Nkg^-1 to 3sf. The mass of the moon is 7.35x10^22kg.

The moons gravitational pull on the earth causes the ocean tides. show that the gravitational pull of the moon on the earth's oceans is approximately 3 millionths of the gravitational pull of the eath on its oceans. Assume the distance from the earth to the moon is 380,000km.

help
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kirino1
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Im not getting anywhere and its really depressing me
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Pangol
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I think this is best approached by working out the Moon's gravitational field strength at the surface of the Earth, and then comparing it to the Earth's gravitational field strength at the surface of the Earth. Do you know how to do that?
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PatrickD
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The AQA specifications (A and B) make use of formulae booklets given to candidates. Take a look at the 'Gravitational fields and Mechanics' section of this booklet (either in the back of your textbook, in a separate booklet or on the AQA website) and compare the formulae to what information you have and what you want to calculate.
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kirino1
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(Original post by Pangol)
I think this is best approached by working out the Moon's gravitational field strength at the surface of the Earth, and then comparing it to the Earth's gravitational field strength at the surface of the Earth. Do you know how to do that?
thanks. I got the answer
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chrisburns123
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is there any chance you could explain this? I've tried to do this by using g= GM/r^2 for both the moon, then dividing the relevant values of g into each other, but I'm getting x10^5 answers, not x10^6 :s
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Pangol
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What value do you get for the gravitational field strength of the Moon at the Earth's surface? Can you show how you get it?
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TheHaylio
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(Original post by kirino1)
thanks. I got the answer

Couldn't find a way to message you! I havent heard back from them yet no, you?
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kirino1
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(Original post by TheHaylio)
Couldn't find a way to message you! I havent heard back from them yet no, you?
Thats odd.. and i havent either... i hope they tell us soon though
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Chicken Bacon
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(Original post by kirino1)
thanks. I got the answer
I'm trying the same question, care to explain how it's done ? i calculated g for the moon on the earths surface, and of course g for the earth on its own oceans is 9.8. but i get

g for moon = 13000
g for earth = 9.8

this is not 3 millionth ?
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Stonebridge
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(Original post by iAre Teh Lejend)
I'm trying the same question, care to explain how it's done ? i calculated g for the moon on the earths surface, and of course g for the earth on its own oceans is 9.8. but i get

g for moon = 13000
g for earth = 9.8

this is not 3 millionth ?
Same question has been asked twice
Here
http://www.thestudentroom.co.uk/show....php?t=2156314

How can anyone know where you have gone wrong if you don't post your solution?
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Yorrap
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(Original post by Stonebridge)
Same question has been asked twice
Here
http://www.thestudentroom.co.uk/show....php?t=2156314

How can anyone know where you have gone wrong if you don't post your solution?
I think question is wrong.
My exam board is Edexcel but since the question is a good practice, I gave it a try;

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Stonebridge
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It's correct.


You have calculated the Moon's field correctly.
If g is 10 on Earth (approx) and the value of the moon's field at the earth is 3 x 10-5
then the ratio of the Moon's field at Earth to the Earth's field at Earth is (3 x 10-5) / 10
Which is 3 x 10-6
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Yorrap
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(Original post by Stonebridge)
It's correct.


You have calculated the Moon's field correctly.
If g is 10 on Earth (approx) and the value of the moon's field at the earth is 3 x 10-5
then the ratio of the Moon's field at Earth to the Earth's field at Earth is (3 x 10-5) / 10
Which is 3 x 10-6

Do I understand the question wrongly?

Hmm :hmmmm:
show that the gravitational pull of the moon on the earth's oceans is approximately 3 millionths of the gravitational pull of the eath on its oceans.
Doesn't this mean;
"show that the field strength of the moon on the Earth's oceans is approximately 1/3000000 of the gravitational field strength of the earth on its oceans?"

(in other words show that gMoon is approximately 9.81/3 000 000 ?)


If g of Moon on Earth's oceans = 3.36x10-5 Nkg-1
and
g of Earth on its oceans = 9.81 Nkg-1

then I would do the following step and expect to get ~3 000 000 as the answer. ;
gEarth / gMoon = ~ 3x106

but 3.36x10-5 / 9.81 Nkg-1 = 291 964.28 ~ 292 000 = 2.92x105 (or 0.292x106)

so gMoon on Earth's oceans is 0.3 millions times less than gEarth on its oceans and not 3 millions.
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Stonebridge
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I've explained what the question is asking for and shown you how the calculation gives the answer they want.
I admit the wording of the question can be misinterpreted.
I'm certain that, in this question, the whole point being tested is that you
- know g on the Earth's surface due to the Earth is approx 10m/s (You did)
- calculate correctly the value of the Moon's field strength at the Earth (you did)
- realize this is a "ball park" figure. (do you?)

I'll say it again.
If the one is 3 x 10-5 (approx) and the other is 10 (approx) then the first one is what the question has described as "about 3 millionths the value of" the second. 3 x 10-6

So the point of the question was that you correctly get that value of approx. 3 x 10-5

Think
Let's say the first value was 3 and the second was 1,000,000
What fraction is this?
Spoiler:
Show

3 millionths ie 3 / 1,000,000


Multiply both top and bottom by 10-5
you get
3 x 10-5 and 10

What is this fraction?

Spoiler:
Show

It hasn't changed. The same fraction as before.


This is the fraction you are working out.
The ratio of the Moon's field to that of the Earth at that point.

Maybe you are getting mixed up between "1 in 3 million" and "3 millionths"? These are not the same.
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Yorrap
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(Original post by Stonebridge)
I've explained what the question is asking for and shown you how the calculation gives the answer they want.
I admit the wording of the question can be misinterpreted.
I'm certain that, in this question, the whole point being tested is that you
- know g on the Earth's surface due to the Earth is approx 10m/s (You did)
- calculate correctly the value of the Moon's field strength at the Earth (you did)
- realize this is a "ball park" figure. (do you?)

I'll say it again.
If the one is 3 x 10-5 (approx) and the other is 10 (approx) then the first one is what the question has described as "about 3 millionths the value of" the second. 3 x 10-6

So the point of the question was that you correctly get that value of approx. 3 x 10-5

Think
Let's say the first value was 3 and the second was 1,000,000
What fraction is this?
Spoiler:
Show

3 millionths ie 3 / 1,000,000


Multiply both top and bottom by 10-5
you get
3 x 10-5 and 10

What is this fraction?

Spoiler:
Show

It hasn't changed. The same fraction as before.


This is the fraction you are working out.
The ratio of the Moon's field to that of the Earth at that point.

Maybe you are getting mixed up between "1 in 3 million" and "3 millionths"? These are not the same.
Thank you!
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JessyJoy
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Moon's pull on oceans:
g=(6.67x10^-11)(7.35x10^22)/(380000x10^3)^2
g=3.4x10^-5

Earth's pull on oceans:
g=(6.67x10^-11)(5.98x10^24)/(6.37x10^6)^2
g=9.8

3.4x10^-5/9.8=3.5x10^-6
Which means 3.4x10^-5 is approximately 3 millionths of 9.8 because 3 millionths = 3/1x10^6 = 3x10^-6
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CurrentDude
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(Original post by JessyJoy)
Moon's pull on oceans:
g=(6.67x10^-11)(7.35x10^22)/(380000x10^3)^2
g=3.4x10^-5

Earth's pull on oceans:
g=(6.67x10^-11)(5.98x10^24)/(6.37x10^6)^2
g=9.8

3.4x10^-5/9.8=3.5x10^-6
Which means 3.4x10^-5 is approximately 3 millionths of 9.8 because 3 millionths = 3/1x10^6 = 3x10^-6
I agree with your answer except for the language.
Instead of saying "Moon's pull on ocean" I'd use, Gravitation field strength of the moon at 38x10^7 is

Moon's g = 6,67x10^-11xmass of moon / (38x10^7)^2 and then follow the steps you've used.
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Absent Agent
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(Original post by zzxxDash53xxzz)
eassssyyyyyyyyy u serious u look dumb
Actually that's not easy! Why do you think that despite moon's gravitational pull on the oceans of earth being 3 millionths of the gravitational pull of the earth on its oceans, it can still cause ocean tides and, while it does, it cannot accelerate the oceans towards itself?
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