F = Gm₁m₂/r². Plug in masses and distance. Why this force causes tides.

Calculate the Gravitational Force Between Earth and Moon — Newton's Law

Question

Calculate the gravitational force of attraction between the Earth and the Moon.

Given:

Mass of Earth, $M_E = 6 \times 10^{24}$ kg
Mass of Moon, $M_M = 7.4 \times 10^{22}$ kg
Distance between Earth and Moon, $r = 3.84 \times 10^8$ m
Universal gravitational constant, $G = 6.67 \times 10^{-11}$ N m² kg⁻²

Solution — Step by Step

The gravitational force between any two masses is:

F = \frac{G m_1 m_2}{r^2}

Here, $m_1$ and $m_2$ are the two masses, and $r$ is the distance between their centres. Not their surfaces — this distinction trips up a lot of students.

G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}

m_1 = M_E = 6 \times 10^{24} \text{ kg}

m_2 = M_M = 7.4 \times 10^{22} \text{ kg}

r = 3.84 \times 10^8 \text{ m}

We need $r^2$ first. Squaring the distance gives us:

r^2 = (3.84 \times 10^8)^2 = 14.75 \times 10^{16} = 1.475 \times 10^{17} \text{ m}^2

Multiply $G$ , $m_1$ , and $m_2$ together:

G \times m_1 \times m_2 = 6.67 \times 10^{-11} \times 6 \times 10^{24} \times 7.4 \times 10^{22}

Multiply the numbers: $6.67 \times 6 \times 7.4 = 6.67 \times 44.4 \approx 296.15$

Multiply the powers of 10: $10^{-11} \times 10^{24} \times 10^{22} = 10^{35}$

So numerator $= 296.15 \times 10^{35} = 2.9615 \times 10^{37}$

F = \frac{2.9615 \times 10^{37}}{1.475 \times 10^{17}}

F = \frac{2.9615}{1.475} \times 10^{37-17} = 2.009 \times 10^{20} \text{ N}

\boxed{F \approx 2.0 \times 10^{20} \text{ N}}

Why This Works

Newton’s Law of Gravitation tells us that every object with mass attracts every other object with mass. The force depends on both masses (directly) and the square of the distance between them (inversely). Double the distance, and the force drops to one-fourth — this is the inverse square law.

The constant $G = 6.67 \times 10^{-11}$ is extremely small, which is why gravity between ordinary objects (your pen and your notebook) is undetectable. Only when at least one mass is astronomically large — like Earth or the Moon — does gravity become the dominant force shaping the motion of objects.

This force of $2 \times 10^{20}$ N is what keeps the Moon in its orbit around Earth. The same interaction is responsible for ocean tides — the Moon pulls the water on Earth’s near side slightly harder than it pulls the solid Earth, causing the familiar bulge.

Alternative Method — Scientific Notation Handling

Some students struggle with managing the powers of 10 when numbers get large. Here’s a cleaner way to organise the calculation:

Group all powers of 10 separately from the significant figures:

F = \frac{(6.67)(6)(7.4)}{1.475} \times \frac{10^{-11} \cdot 10^{24} \cdot 10^{22}}{10^{17}}

Numerical part: $\frac{6.67 \times 44.4}{1.475} = \frac{296.15}{1.475} \approx 200.8$

Powers of 10: $10^{(-11+24+22-17)} = 10^{18}$

F = 200.8 \times 10^{18} = 2.008 \times 10^{20} \text{ N}

Same answer, and no intermediate step where you can accidentally mis-count the zeros.

In CBSE board exams, you get marks for writing the formula, substituting values, and final answer with units. Even if your arithmetic slips, you can score 2 out of 3 marks by showing the correct setup. Always write the formula first.

Common Mistake

The most common error: using $r$ as the distance from Earth’s surface to the Moon’s surface, rather than centre to centre. The correct distance is always measured between the centres of the two bodies. In this problem, $3.84 \times 10^8$ m is already the centre-to-centre distance, so you use it directly. But if a question gives you surface-to-surface distance, you must add both radii before substituting into the formula.

A related slip: squaring only the coefficient and forgetting to square the power of 10. $(3.84 \times 10^8)^2$ is $14.75 \times 10^{16}$ , not $14.75 \times 10^8$ . When you square a number in scientific notation, both the coefficient AND the exponent get affected — the exponent doubles.

Question

Solution — Step by Step

Why This Works

Alternative Method — Scientific Notation Handling

Common Mistake

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