Calculating the Distance of L1 From Earth

  • Thread starter Thread starter StephenPrivitera
  • Start date Start date
  • Tags Tags
    Earth
AI Thread Summary
The distance of the L1 point from Earth can be estimated using the balance of gravitational forces from the Sun and Earth. At L1, the centripetal force is zero, leading to the equation F_sun = F_earth. By simplifying the equation and neglecting the mass of the object at L1, the relationship can be expressed as (GM_sun)/(r^2) = (GM_earth)/(a-r)^2. Rearranging this yields an estimate for r, which is the distance from the object to the Sun. This calculation provides an approximation, acknowledging that actual distances may vary due to orbital eccentricities and positional changes.
StephenPrivitera
Messages
360
Reaction score
0
Estimate the distance of the L1 point from Earth.

S-------------r------------L1--(a-r)--E

The centripetal force on this object at L1 is equal to the net force on the object. At L1, (to the left of L1 is negative)
F_net=F_sun+F_earth
-(mv2)/r=-GMsm/r2+GMem/(a-r)2
Solving for r gives a polynomial of 5th degree!
How do I go about this in an easier fashion?
 
Physics news on Phys.org
The key word is your first one: Estimate. Make use of the fact that the sun's mass is far larger than the earth's. This will mean for example that a-r<<a. Rewrite the equation in terms of say a and delta = a-r. Then you will find terms like (a-delta)^3, which can be approximated as a^3-3a^2*(delta). After that, it's easy.
 


Calculating the distance of L1 from Earth can be done using the formula for centripetal force. At L1, the force of gravity from the sun and the force of gravity from Earth are equal and opposite, creating a net force of zero. This can be written as: F_net = F_sun + F_earth. Since we are looking for the distance, we can set the two forces equal to each other and solve for r.

The equation will look like this:

F_sun = F_earth

(GM_sun*m)/(r^2) = (GM_earth*m)/(a-r)^2

where G is the gravitational constant, M_sun and M_earth are the masses of the sun and Earth respectively, m is the mass of the object at L1, r is the distance between the object and the sun, and a is the distance between the object and Earth.

To make the calculation easier, we can use the fact that the mass of the object at L1 is very small compared to the masses of the sun and Earth. This means we can neglect the mass of the object in the equation, leaving us with:

(GM_sun)/(r^2) = (GM_earth)/(a-r)^2

We can rearrange this equation to solve for r:

r = [(a^2*GM_sun)/(a^2-GM_earth)]^(1/3)

Plugging in the values for the masses and distances, we can estimate the distance of L1 from Earth. Keep in mind that this is just an estimate, as the actual distance can vary due to factors such as the eccentricity of Earth's orbit and the changing positions of the sun and Earth.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top