The Sun and Earth each exert a gravitational force on the Moon. What is the ratio FSun/FEarth of these two forces? (The average Sun-Moon distance is equal to the Sun-Earth distance.)
The ratio of the forces is
Fsun/Fearth
= (Msun/Mearth)*Dearth/Dsun)^2
Dearth is the average earth-moon distance and Dsun is the average moon-sun (or Earth-sun) distance.
The relative influence upon tides is different from this number. It depends upon the gradient, not the force itself.
To find the ratio FSun/FEarth, we need to use Newton's law of universal gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Let's denote the mass of the Sun as MSun, the mass of the Earth as MEarth, and the distance between the Sun and the Earth (or the average Sun-Moon distance) as D.
According to the problem statement, the Sun-Moon distance is equal to the Sun-Earth distance, so we can write:
D(Sun-Earth) = D(Sun-Moon)
Now, let's calculate the ratio FSun/FEarth:
FSun = G * MSun * MMoon / D^2
FEarth = G * MEarth * MMoon / D^2
Dividing FSun by FEarth:
FSun/FEarth = G * MSun * MMoon / D^2 / (G * MEarth * MMoon / D^2)
= (G/G) * (MSun/MEarth) * (MMoon/MMoon) * (D^2/D^2)
= MSun / MEarth
Therefore, the ratio FSun/FEarth is simply equal to the ratio of the masses of the Sun and the Earth.
To find the ratio FSun/FEarth, we need to calculate the gravitational forces exerted by the Sun and Earth on the Moon and then divide them. The gravitational force between two objects can be calculated using Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2), m1 and m2 are the masses of the two objects, and r is the distance between them.
Given that the average Sun-Moon distance is equal to the Sun-Earth distance, the distances cancel out in the ratio FSun/FEarth. Therefore, we only need to consider the masses of the Sun and Earth.
Let's say the mass of the Sun is MSun and the mass of the Earth is MEarth.
The gravitational force exerted by the Sun on the Moon (FSun) is given by:
FSun = G * (MSun * MMoon) / (RSun-Moon)^2
where MMoon is the mass of the Moon and RSun-Moon is the distance between the Sun and the Moon.
Similarly, the gravitational force exerted by the Earth on the Moon (FEarth) is given by:
FEarth = G * (MEarth * MMoon) / (REarth-Moon)^2
where REarth-Moon is the distance between the Earth and the Moon.
Since the average Sun-Moon distance is equal to the Sun-Earth distance, RSun-Moon is the same as REarth-Moon. Thus, we can conclude that the ratio FSun/FEarth is simply:
FSun/FEarth = (MSun * MMoon) / (MEarth * MMoon)
The mass of the Moon (MMoon) cancels out of the equation, resulting in:
FSun/FEarth = MSun / MEarth
Therefore, the ratio of the gravitational forces FSun/FEarth between the Sun and the Moon is equal to the ratio of their masses.