# AQA Physics A2 HELP

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#1
This question is driving me insane.

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

Question:
The moon's 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 Earth on its oceans. Assume that the distance from the Earth to the Moon is 380000km.

What is meant by 'gravitational pull'

Assuming they mean Gravitational field strength, this is my working out:

Gravitational field strength, g, of the earth on its oceans:

g = 9.8Nkg^-1 ( of course, no need for calculations )

Gravitational field strength, g, of the moon on the earths oceans:

g = GM / r^2
g = 6.67x10^-11 x 7.35x10^22 / 380000000
g = 12901.18421 = 13000Nkg^-1

The ratio definitely isn't 3million, where did I go wrong ? help please
0
8 years ago
#2
(Original post by iAre Teh Lejend)
This question is driving me insane.

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

Question:
The moon's 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 Earth on its oceans. Assume that the distance from the Earth to the Moon is 380000km.

What is meant by 'gravitational pull'

Assuming they mean Gravitational field strength, this is my working out:

Gravitational field strength, g, of the earth on its oceans:

g = 9.8Nkg^-1 ( of course, no need for calculations )

Gravitational field strength, g, of the moon on the earths oceans:

g = GM / r^2
g = 6.67x10^-11 x 7.35x10^22 / 380000000
g = 12901.18421 = 13000Nkg^-1

The ratio definitely isn't 3million, where did I go wrong ? help please
Same question here
Does this help?
http://www.thestudentroom.co.uk/show...php?p=39631119
0
#3
(Original post by Stonebridge)
Same question here
Does this help?
http://www.thestudentroom.co.uk/show...php?p=39631119
Thanks for your reply, but i am still confused. In my posted method, I did what you suggested, didn't I ?
0
8 years ago
#4
(Original post by iAre Teh Lejend)
Thanks for your reply, but i am still confused. In my posted method, I did what you suggested, didn't I ?
On looking closer...
There's an error in the calculation not the method.
You forgot to square the distance of the Moon to the Earth.
0
4 years ago
#5
Hi!
Question.. why don't we include the radius of the moon in this? Its probably obvious but I'm missing the point....

Surely when we work out g on the surface of the earth we use the radius, so g on the surface of the moon would use the radius.. then include the distance?

g(moon) = Gx(7.35E22) / (380 000 000 X 1740E3)^2 (since the question earlier gives us the radius of the moon?)

Thanks!
James
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