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# Gravitational attraction watch

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1. A body is placed between the moon and the earth in a position where it feels equal attraction to both bodies. How far is it from the earth?(The radius of the moons orbit is 3.8x10^8)

What does it mean equal attraction? Is this where the field strength at the body is equal to that from the moon and the earth? ie. Gm(earth)/r(earth)^2 = Gm(moon)/r(moon)^2

???
2. (Original post by imasillynarb)
A body is placed between the moon and the earth in a position where it feels equal attraction to both bodies. How far is it from the earth?(The radius of the moons orbit is 3.8x10^8)

What does it mean equal attraction? Is this where the field strength at the body is equal to that from the moon and the earth? ie. Gm(earth)/r(earth)^2 = Gm(moon)/r(moon)^2

???
It means the gravitational force between the Earth and the body is equal to the force between the Moon and the body:

F = -GMm/r^2

Gm1m2/(r1)^2 = Gm1m3/(r2)^2

where m1 = mass of the body, m2 = mass of the Earth, m3 = mass of the Moon, r1 = distance between Moon and body and r2 = distance between Earth and body.

I'm not sure why they've given you the radius of the Moon, unless they want you to take the force as acting from the centre of the Moon (but then they'd have to give you the radius of the body )
3. (Original post by imasillynarb)
A body is placed between the moon and the earth in a position where it feels equal attraction to both bodies. How far is it from the earth?(The radius of the moons orbit is 3.8x10^8)

What does it mean equal attraction? Is this where the field strength at the body is equal to that from the moon and the earth? ie. Gm(earth)/r(earth)^2 = Gm(moon)/r(moon)^2

???
It means that the force of the body's weight due to the Earth is completely cancellled by the force of the body's weight due to the moon. So if you fixed the Earth and the moon in that position (by some huge clamp), the body would have no resultant force at this point.

This point must be (a) on the line linking the Earth and moon's centres of mass (because otherwise both would cause a "horizontal" component of resultant force and accelerate the body towards this line), and (b) where the two forces are equal:

(b) = body, (m) = moon, (e) = Earth:

F(mb) = Gm(b)m(m)/r(mb)²
F(eb) = Gm(E)m(b)/r(eb)²

Equate these two, as you want F(mb) and F(eb) to be equal:

Gm(b)m(m)/r(mb)² = Gm(e)m(b)/r(eb)²
m(m)r(eb)² = m(e)r(mb)²

Now the radius of the moon's orbit (assume to be the distance between the Earth and the moon) is 3.8x10^8 (metres I hope):

So:

r(eb) + r(mb) = 3.8 x 10^8
r(mb) = 3.8 x 10^8 - r(eb)

We also know (from google) that:

m(e) = 5.97 × 10^24 kg
m(m) = 7.35 x 10^22 kg

We are trying to find r(eb), so plugging in numbers:

m(m)r(eb)² = m(e)r(mb)²
(7.35 x 10^22)(r(eb))² = (5.97 × 10^24)(3.8 x 10^8 - r(eb))²
(7.35 x 10^22)(r(eb))² = (5.97 × 10^24)(1.44 x 10^17 + r(eb)² - 7.6 x 10^8 x r(eb))
7.35 x 10^22 x r(eb)² - (5.97 × 10^24)(1.44 x 10^17 + r(eb)² - 7.6 x 10^8 x r(eb)) = 0
7.35 x 10^22 x r(eb)² - 8.62 x 10^41 - 5.97 x 10^24 x r(eb)² + 4.54 x 10^33 x r(eb)) = 0
-5.90 x 10^24 r(eb)² + 4.54 x 10^33 x r(eb)) - 8.62 x 10^41 = 0

a = -5.90 x 10^24
b = 4.54 x 10^33
c = -8.62 x 10^41

r(eb) = -b ± √(b²-4ac) /2a
r(eb) = -4.54 x 10^33 ± √((4.54 x 10^33)²-4(-5.90 x 10^24)(-8.62 x 10^41))/2(-5.90 x 10^24)
r(eb) = -4.54 x 10^33 ± √(1.55 x 10^67)/(-1.18 x 10^25)
r(eb) = 5.11 x 10^7 m OR 7.18 x 10^8 m

I'm guessing the latter is when the body is closer to the moon.

Hmm... I've just worked out the distance where the weight of the body is equal and in the same direction. Tried to correct by making the RHS negative but got no real roots, and there definately is a point where the body would remain at rest. :s

Heh.
4. (Original post by Nylex)
I'm not sure why they've given you the radius of the Moon, unless they want you to take the force as acting from the centre of the Moon (but then they'd have to give you the radius of the body )
I think it was the radius of the Moon's orbit, not of the actual moon
5. mik1a - i don't think there's any time in physics a-level you'd need the quadratic formula - always a sign you've gone wrong!

(7.35*10^22)(r(eb))² = (5.97*10^24)(3.8*10^8 - r(eb))²
All of these will be positive numbers so you can just take the square root of each side to give
sqrt(7.35*10^22)*r(eb) = sqrt(5.97*10^24)*(3.8*10^8 - r(eb))
r(eb) = sqrt[(5.97*10^24)/(7.35*10^22)]*(3.8*10^8 - r(eb))
r(eb)(1 + 9.012..) = 9.012..*3.8*10^8
r(eb) = 9.012..*3.8*10^8/(1 + 9.012..) = 3.42*10^8 m
6. yeah - i hate quadratics!! theres no way you would need them in a physics paper!
7. (Original post by Bezza)
mik1a - i don't think there's any time in physics a-level you'd need the quadratic formula - always a sign you've gone wrong!

(7.35*10^22)(r(eb))² = (5.97*10^24)(3.8*10^8 - r(eb))²
All of these will be positive numbers so you can just take the square root of each side to give
sqrt(7.35*10^22)*r(eb) = sqrt(5.97*10^24)*(3.8*10^8 - r(eb))
r(eb) = sqrt[(5.97*10^24)/(7.35*10^22)]*(3.8*10^8 - r(eb))
r(eb)(1 + 9.012..) = 9.012..*3.8*10^8
r(eb) = 9.012..*3.8*10^8/(1 + 9.012..) = 3.42*10^8 m
Haha... so much simpler!

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