Calculate the distance between the center of the sun at which particle experiences equal attractions from the earth an the sun?

Answer 1

OK. Let's use the following symbols:

M = mass of the sun

P = mass of the particle

E = mass of the earth

G = gravitational constant

Rs = distance from the sun

Re = distance from the earth

Fs = force between the particle and the sun

Fe = force between the particle and the earth

From Newton:

Fs = G (M P) / (Rs)2

Fe = G (E P) / (Re)2

We want to know where these forces are equal:

G (M P) / (Rs)2 = G (E P) / (Re)2

M / (Rs)2 = E / (Re)2

(Rs/Re)2 = M/E

Rs/Re = sqrt(M/E)

This gives us the ratio of distances to the sun and earth, and we can worry about the

actual miles later. Just now, we only need the square-root of the ratio of the masses.

sqrt(M/E) = sqrt[ (1.98X1030) / (5.97x1024) ] = sqrt(331658.29) = 575.9 (rounded)

Now we're ready to do the actual miles. Notice that the particle does not have to be on

the straight line between the earth and sun, as long as the ratio of the two distances is

correct. Just to keep it simple, though, we'll place the particle on the earth-sun line.

Distance between the earth and sun: 93 million miles

Ratio of the distances to earth and sun from the particle: 1 to 575.9

Distance from the earth: (93 million) x (1/576.9) = 161,207

Distance from the sun: (93 million) x (575.9/576.9) = 92,838,794

One very interesting thing to notice:

The position for equal attraction by earth and sun, on the line joining them, is about

161,000 miles from earth, or about 1/576th of the distance.

The moon's average distance from the earth is 238,000 miles, or about 1/391 of the

sun's distance.

So for a considerable portion of the moon's orbit, the force of the sun's attraction on it

exceeds the force of the earth's attraction ! Verrrrry interesting . . .