Where Is the Point of Zero Net Gravitational Force Between Earth and the Moon

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SUMMARY

The point of zero net gravitational force between the Earth and the Moon is located approximately 3.5 x 10^5 meters from the Earth's center. This conclusion is derived from the gravitational force equation Fg = Gm1m2/R^2, where the mass of the Moon is 1.2% of the Earth's mass. The discussion highlights the importance of correctly defining variables and converting units, as the textbook answer may have been in kilometers rather than meters. Participants confirmed that the mass of the object can be eliminated from the equations since it appears on both sides.

PREREQUISITES
  • Understanding of gravitational force equations, specifically Fg = Gm1m2/R^2
  • Knowledge of basic algebra for rearranging equations and solving quadratics
  • Familiarity with unit conversions between meters and kilometers
  • Concept of gravitational interactions between two bodies
NEXT STEPS
  • Learn how to derive gravitational force equations in different contexts
  • Study the concept of gravitational equilibrium points in multi-body systems
  • Explore the implications of mass ratios in gravitational calculations
  • Investigate the effects of distance on gravitational force strength
USEFUL FOR

Students studying physics, particularly those focusing on gravitational forces, as well as educators seeking to clarify concepts related to gravitational equilibrium between celestial bodies.

elasticities
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Homework Statement


At a certain point between Earth and the Moon, the net gravitational force exerted on an object by the Earth and the Moon is ZERO. The mass of the Moon is 1.2% the Mass of the Earth. The centre to centre distance between the Moon and the Earth is 3.84*10^5 km.

i) WHERE IS THIS POINT LOCATED?
ii) What is the meaning of the quadratic root whose value exceeds the Earth-Moon distance?

Homework Equations


Fg=Gm1m2/R^2

The Attempt at a Solution


Fnet=0
Fnet=Fmoon-Fearth
Fmoon=Fearth
Rmoon=x
Rearth=3.84*10^8m-x
Mmoon=1.2
Mearth=100

Uh...what's next? And am I right so far?
 
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So far, so good. Now use the Gravity equation to get expressions for Fmoon and Fearth. Hint: Let the mass of the object be 'm'.
 
Doc Al said:
So far, so good. Now use the Gravity equation to get expressions for Fmoon and Fearth. Hint: Let the mass of the object be 'm'.

Do I need to put a value in for 'm' or can I just get rid of it since its on both sides of the equation.
 
elasticities said:
Do I need to put a value in for 'm' or can I just get rid of it since its on both sides of the equation.
What do you think? :wink:
 
100/(3.84*10^8m-x)^2 = 1.2/x^2

Is that right? Do I isolate for x?
 
elasticities said:
100/(3.84*10^8m-x)^2 = 1.2/x^2

Is that right? Do I isolate for x?
Looks good to me. You'll have to solve for x. Rearrange terms to put the quadratic into standard form.
 
Doc Al said:
Looks good to me. You'll have to solve for x. Rearrange terms to put the quadratic into standard form.

I'm not getting the right answer which is supposed to be 3.5*10^5m from Earth's centre. I will try again.
 
elasticities said:
I'm not getting the right answer which is supposed to be 3.5*10^5m from Earth's centre. I will try again.
Realize that you've defined your variable X to be the distance from the Moon's center. Once you have X, you can then figure out the distance from the Earth's center.
 
Ok thanks, I think the textbook answer was just wrong. :)
 
  • #10
elasticities said:
Ok thanks, I think the textbook answer was just wrong. :)
I suspect that the book's answer was in km, not m.
 

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