- #1
etotheipi
- Homework Statement
- Derive expressions for the gravitational field due to the moon at the near and far sides of the Earth, and use these to estimate the height ##h## of the tidal bulge of water. Assume the Earth is covered with a layer of water 1000m deep.
- Relevant Equations
- ##g = \frac{GM}{r^{2}}##
I'm having some conceptual difficulty with this; here's what (little!) I've done so far.
Suppose the distance between the centres of the Earth and the moon is ##x## and that the radius of the Earth is ##r##, and let the gravitational field strength due to the moon at the near side, far side and centre of the Earth be ##g_{a}##, ##g_{b}## and ##g_{c}## respectively. Then
##g_{a} = \frac{GM_{m}}{(x-r)^{2}} \approx \frac{GM_{m}}{x^{2}}(1+\frac{2r}{x}) = g_{c} + \frac{2GM_{m}r}{x^{3}}##
and
##g_{b} = \frac{GM_{m}}{(x+r)^{2}} \approx \frac{GM_{m}}{x^{2}}(1-\frac{2r}{x}) = g_{c} - \frac{2GM_{m}r}{x^{3}}##
I'm at a loss for what to do next. I considered transforming into a rotating frame centred on the moon and imagined a mass element ##dm## at each bulge which would have a fictitous ##dm(x-r)\omega^{2}## acting (assuming ##h\ll r##) in addition to ##dmg_{a}## and the weight due to the Earth at a height ##h## above the surface, which I obtained to be ##dmg_{h} = dm(g_{E} -\frac{2GM_{E}h}{r^{3}})##. However, when I did the force balance and substituted in the known constants (plus taking ##\omega = \frac{2\pi}{(28)(24)(60)(60)}##), I got an answer for ##h## which was many orders of magnitude greater than common sense would suggest.
I then thought to see if I could work out the effective gravitational potential energy at the surface of the bulge and equate this to that at the poles (taking the surface of the water to be an equipotential). This threw out, when assuming the depth at the poles is 1000m,
$$h(g_{E} - \frac{GM_{m}}{x^{2}}(1+\frac{2r}{x})) = 1000g_{E} \implies h = 1000.0005m$$
which is only ##0.0005## metres higher than at the poles. So this has to be wrong as well.
I was wondering if someone point me in the right direction, thanks!
Suppose the distance between the centres of the Earth and the moon is ##x## and that the radius of the Earth is ##r##, and let the gravitational field strength due to the moon at the near side, far side and centre of the Earth be ##g_{a}##, ##g_{b}## and ##g_{c}## respectively. Then
##g_{a} = \frac{GM_{m}}{(x-r)^{2}} \approx \frac{GM_{m}}{x^{2}}(1+\frac{2r}{x}) = g_{c} + \frac{2GM_{m}r}{x^{3}}##
and
##g_{b} = \frac{GM_{m}}{(x+r)^{2}} \approx \frac{GM_{m}}{x^{2}}(1-\frac{2r}{x}) = g_{c} - \frac{2GM_{m}r}{x^{3}}##
I'm at a loss for what to do next. I considered transforming into a rotating frame centred on the moon and imagined a mass element ##dm## at each bulge which would have a fictitous ##dm(x-r)\omega^{2}## acting (assuming ##h\ll r##) in addition to ##dmg_{a}## and the weight due to the Earth at a height ##h## above the surface, which I obtained to be ##dmg_{h} = dm(g_{E} -\frac{2GM_{E}h}{r^{3}})##. However, when I did the force balance and substituted in the known constants (plus taking ##\omega = \frac{2\pi}{(28)(24)(60)(60)}##), I got an answer for ##h## which was many orders of magnitude greater than common sense would suggest.
I then thought to see if I could work out the effective gravitational potential energy at the surface of the bulge and equate this to that at the poles (taking the surface of the water to be an equipotential). This threw out, when assuming the depth at the poles is 1000m,
$$h(g_{E} - \frac{GM_{m}}{x^{2}}(1+\frac{2r}{x})) = 1000g_{E} \implies h = 1000.0005m$$
which is only ##0.0005## metres higher than at the poles. So this has to be wrong as well.
I was wondering if someone point me in the right direction, thanks!
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