Fractional Weight Reduction from Gravitational Attraction of Water Tank

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SUMMARY

The discussion centers on calculating the fractional weight reduction experienced by a person standing 17 meters below a spherical water tank containing 6.0×106 kg of water due to gravitational attraction. Participants utilize Newton's law of gravitation, expressed as F = GMm/r2, to derive the forces exerted by both the water tank and the Earth. The key conclusion is that the ratio of these forces provides the fraction of weight reduction, which is calculated as ΔF/F = (F_t/F_e), where F_t is the gravitational force from the tank and F_e is the gravitational force from the Earth.

PREREQUISITES
  • Understanding of Newton's law of gravitation (F = GMm/r2)
  • Familiarity with gravitational force calculations
  • Basic knowledge of ratios and fractions in physics
  • Concept of weight as a measure of gravitational force
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  • Study gravitational force calculations in detail using different mass scenarios
  • Explore the implications of Gauss's law in gravitational fields
  • Learn about the concept of weight reduction in varying gravitational fields
  • Investigate the effects of nearby massive objects on gravitational attraction
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Students in physics, educators teaching gravitational concepts, and anyone interested in understanding the effects of gravitational forces on weight measurements.

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Homework Statement



If you're standing on the ground 17 m directly below the center of a spherical water tank containing 6.0×106kg of water, by what fraction is your weight reduced due to the gravitational attraction of the water?


Homework Equations



F = GMm/r^2

The Attempt at a Solution



not sure where to start.
 
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anyone?
 
So that equation gives you the gravitational attraction force. You can use it to calculate the force between u and the water tank, between u and the earth, etc.
The tank and Earth are both spheres, so u can use Gauss's law and symmetry..
 
Still a little confused.

I get I need to calculate the force between the tank and I . but then I am confused on where to go.
 
You can calculate the force between you and the water tank, then do the same thing between you and the Earth (which should essentially be F=ma) and compare. The centre of the water tank directly opposes the centre of the Earth so they are opposite forces. The effect of the water tank will be very minor, but still interesting to calculate. If what I said does not make sense let me know and I will try to reword it all. Also, weight is a measure of force, so by comparing the two forces you can solve.

I don't have my mechanics text near me for the constants, but if you calculate the force exerted by the tank and subtract it from the force exerted by the Earth then the resulting force is your weight in Newtons
 
so,

F_g_t = GMm/r^2

= G(6.0*10^6) M_p / 17^2

and

F_g_e = GMm/r^2

=

G(5.98*10^24)M_p / (6378.1)^2

is this the correct step so far?
If so then what should I do next?
 
It might have been easier to solve it as variables first.

For instance they want the ratio of the forces, so ...

F/Fo = (Gm(M_tank)/d2)/Gm(M_earth)/R_earth2)

Simplifying you have

Ratio = M_tank*R_earth2/M_earth*d2

Then just calculate.
 
Last edited:
M_tank = 6.0 *10^6
M_earth = 5.98 * 10^24
d = 17m
r_e = 6.37 * 10^6

my answer came out as 5.58 * 10^-8
But I guess this is wrong
 
forget it this is solved. I miscalculated. Thanks
 
  • #10
1 question. Why is it the force of Earth over the force of the spherical tank?

Why does that give you the fraction of you weight reduced?
 
  • #11
tnutty said:
1 question. Why is it the force of Earth over the force of the spherical tank?

Why does that give you the fraction of you weight reduced?

Look again. The Mass of the Tank is in the numerator, Earth below. Do the algebra for dividing the expressions for gravity and you will see.

As to the fraction ...

they want ΔF / F

ΔF = Ft
 
Last edited:
  • #12
alright thanks, but I am not sure how f1/f2 gives you the fraction of your reduced weight.

Although I get the math, I don't get the concept of this problem. Would you
mind explaining why a force over another force gives you a fraction of one's reduced weight?

F_t = (m_t)(a_t)

F_e = (m_e)(a_e)

(m_t)(a_t)
------------
(m_e)(a_t)

I get that this gives you a fraction. And that this fraction gives you how
much you weight relative to the Earth ...something?

but how does it tell you how much less you weight.

Can you add some context to the math that you did above?
 
  • #14
I am sorry I am still not getting it.
 
  • #15
Consider you are measuring some mass m's weight.

F = m*g

Well that is also G*M*m/r2

That's weight.

Now they also set you underneath the old town water tank. That has a prodigious mass. And as we know masses have attractiveness to each other and that attractiveness in this case is up.

So ... if it is going up that means it is reducing the weight measured down.

What is the difference in that weight as a fraction of the weight? Well I'd say it's the weight (force) up divided by the weight down. The weight up is your difference in weight, and the weight down is as before nothing changed from weighing the mass in an open flat parking lot.
 
  • #16
OH, nice. Thanks a lot. You know I don't just want to memorize formula's. I want to understand it so that's why I ask. I am thankful for you explanation.
 

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