Relative strength of solar and lunar tides – choose the closest ratio What is the approximate ratio of the solar tidal force to the lunar tidal force acting on the Earth?

Introduction / Context:
Tidal forces arise from differential gravitational attraction across the Earth. Although the Sun is vastly more massive than the Moon, the Moon is much closer; tidal force scales with mass divided by distance cubed, so proximity dominates.

Given Data / Assumptions:

• Tidal force F ∝ M / R^3 for a perturbing body of mass M at distance R.
• Use mean values for the Sun and Moon to obtain an order-of-magnitude ratio.

Concept / Approach:
The ratio (solar/lunar) ≈ (M_sun / R_sun^3) / (M_moon / R_moon^3). Substituting average astronomical values yields a ratio close to 0.46. Therefore the Moon raises tides about twice as effectively as the Sun; spring tides occur when their effects reinforce each other.

Step-by-Step Solution:
Write ratio: (F_sun/F_moon) = (M_sun / R_sun^3) * (R_moon^3 / M_moon).Insert mean values (conceptually): enormous M_sun but much larger R_sun drives the ratio below 1.The accepted practical value is ≈ 0.46 (about 46%).

Verification / Alternative check:
This value is consistent with observed spring–neap tide amplitudes and with standard geophysics texts; the solar tide is roughly 45–50% of the lunar tide.

Why Other Options Are Wrong:

• 0.23 underestimates the solar contribution.
• 0.75 and 1.00 overestimate; the Sun’s larger distance reduces its tidal effect below the Moon’s.
• 2.20 is physically implausible (would imply solar tide bigger than lunar).

Common Pitfalls:
Confusing force proportionality (1/R^2) with tidal force proportionality (1/R^3); using mass only without distance-cubed scaling.

Final Answer:
Approximately 0.46