# Distance and Mass of the Sun from Tidal Forces

We can get the distance to the sun, the mass of the sun, the location of the barycenter and the orbital period of the earth only from the tidal forces of the sun acting on earth.

Measuring the tidal forces of the sun on earth using gravimeters to enough accuracy is very difficult, not because the gravimeters are not accurate enough, but because the tidal forces of the moon interact with the tidal forces of the sun and have to be canceled. Another difficulty is that we need the difference of the tidal forces for our calculations. This difference is even much smaller than the tidal forces themselves.

To get the most effect we have to choose the tidal force nearest to the sun, i.e. the position where the sun is directly overhead, and the farthest distance on the other side of the earth. To check the correctness of the formulas on this page I calculate the expected tidal forces for the already known dimensions of the solar system. I assume this values were measured somehow. In practice the method described here is not feasable and there are much better ways to get the parameters.

The goal of this page is to show that in principle the distance and mass of the sun can be calculated from tidal forces, i.e. from measurements made on earth without any reference to the outside. By calculating the barycenter we can even conclude, that the earth orbits the sun and not the other way around and we can calculate the orbital period of the earth, i.e. how many days a year has.

So lets start:

## Calculation of the Tidal Forces

 (1)
(2)
 (3)
(4)
where'
 $F_n$ ' =' 'tidal force on the near side to the sun $F_f$ ' =' 'tidal force on the far opposite side to the sun $d$ ' =' 'distance to the sun $R$ ' =' 'radius of the earth $M$ ' =' 'mass of the sun $G$ ' =' 'gravitational constant

## Calculating the Distance to the Sun

We assume we have measured the tidal forces $F_n$ and $F_f$. In the equations above we have the 2 unknowns d = distance to the sun and M = mass of the sun. But because we know 2 different tidal forces caused by the sun, we have 2 equations to solve for the distance to the sun, by dividing (2) by (4):

 (5)

Dividing nominator and denominator by $R^3$, to express (5) by the relation d/R:

 (6)

Lets substitude A = d/R, so that if dR then it is 1/A1:

 (7)

Dividing nominator and denominator by A3 yields:

 (8)

Because dR, so that 1/A1, we can neglegt powers of 1/A and get the approximation:

(9)

Now that we got rid of the powers of A we can easily solve for A:

 (10) (11) (12)

And we finally can solve for the distance to the sun:

(13)
for
where'
 $d$ ' =' 'distance to the sun $R$ ' =' 'radius of the earth $F_n$ ' =' 'tidal force on the near side to the sun $F_f$ ' =' 'tidal force on the far opposite side to the sun

Lets put in the values:

The expected values for the tidal forces are calculated with the Calculator for Tidal Forces.

• Radius of the earh: R = 6,371,000 m
• Tidal force per kg (acceleration) on the GP of the sun, expected Fn = 5.0508623 × 10−7 N/kg
• Tidal force per kg (acceleration) on the opposite side, expected Ff = 5.0502217 × 10−7 N/kg

d = 150.7 × 109 m, (accepted value is 149.6 × 109 m)

Note: to get an accuracy of 3 significant digits for d we have to measure the tidal forces to at least 6 significant digits, because we have to calculate the differnce between Fn and Ff and this should give at least 3 digits accuracy. Such accurate measurements are very difficult to achieve in practice for such small forces, if at all. But in principle we can make the calculations from such measurements.

## Calculating the Mass of the Sun

As we now can calculate the distance d to the sun, we can solve (2) for M to get the mass of the sun:

 (14) (15) (16)

To simplify this equation I want to apply the substitution A = R/d. To do so I factor out d4 in the nominator and d2 in the denominator:

Because dR it is A = R/d1 and A2A. So we can make an approximation and get rid of the A2 terms:

 (17)

Now 1/(2A) ≫ 1, so we can neglegt the −1 and simplify further:

 (18)

So the final result is:

(19)
for

Lets use some values:

• Tidal force per kg (acceleration) on the GP of the sun: Fn = 5.0508623 × 10−7 N/kg
• Distance from the sun calculated above: d = 150.7 × 109 m
• Radius of the earh: R = 6,371,000 m
• Gravitational constant: G = 6.674 × 10−11 N·m2/kg2

M = 2.0327 × 1030 kg, (accepted value is 1.9885 × 1030 kg)

That is only 2.2% too much!