Use the law of universal gravitation and the idea of gravitational field strength to calculate forces, g and net force between massive bodies

Learning goal: State the force law for gravitational interaction, F = Gm₁m₂/r², with the meaning and conditions of each symbol; explain gravitational field strength as force per unit mass, g = F/m, and use it to explain why all objects near Earth's surface fall with the same acceleration; calculate g from Earth's mass and radius; find the net force on an object between two massive bodies; and explain why gravity between everyday objects is negligible while a planet's gravity is not.

A universal interaction. Every object with mass attracts every other object with mass. The force is weak: two people standing a meter apart pull on each other gravitationally, but with a force far too small to feel. Gravitational interactions are hard to observe unless at least one of the objects is extremely massive, such as a planet, a moon or a star; the pull between two laboratory-sized masses can be measured, but only with a very sensitive instrument such as a torsion balance, which is how G itself is found. The law that describes the interaction is the same for an apple and the Earth, for the Earth and the Moon, and for two galaxies: the strength of the gravitational force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers,

F = G m₁ m₂ / r².

Here G is the universal gravitational constant, 6.674 × 10⁻¹¹ N·m²/kg² (equivalently m³/(kg·s²)), a number measured in the laboratory that is the same everywhere; m₁ and m₂ are the two masses in kilograms; and r is the distance between the centers of the two objects in meters, not the gap between their surfaces. The law is exact for point masses and for spherical bodies such as planets that do not overlap; for irregular everyday objects close together, using center-to-center distance gives only an estimate. The force on each object has the same magnitude and points toward the other object, a third-law pair. The inverse square is the part students most often misapply: doubling r divides the force by 4, tripling it divides by 9, and halving it multiplies by 4.

Why two people do not notice. Two 70 kg people 1.0 m apart: F = 6.674 × 10⁻¹¹ × 70 × 70 ÷ 1.0² = 3.3 × 10⁻⁷ N. Each person's weight is about 686 N, two billion times larger. The interaction exists; it is simply swamped by every other force in ordinary life. Change one mass to Earth's, 5.97 × 10²⁴ kg, and the distance to Earth's radius, 6.37 × 10⁶ m (both approximate values), and the same law gives the familiar weight.

Gravitational field strength. Rather than recompute the force for every object near Earth, physicists describe Earth's influence at a location by its gravitational field strength: the gravitational force per unit mass that any object placed there would experience, g = F_g/m, in newtons per kilogram. The field belongs to the source and exists whether or not anything is there to feel it; placing a mass in the field only reveals it. The field direction is toward the center of the source. A field diagram shows arrows at a series of locations, each pointing toward the center and drawn longer where the field is stronger; the arrows shorten with distance according to the inverse square. Field line diagrams are not used in this course. From the force law, the field strength of a source of mass M at distance r from its center is g = GM/r². Given g at a location, the force on any mass m placed there is F_g = mg, and a stronger field means a larger force on the same mass.

A fully worked model: g at Earth's surface. g = GM/R² = 6.674 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ (6.37 × 10⁶)². The numerator is 3.98 × 10¹⁴ N·m²/kg; the denominator is 4.06 × 10¹³ m². The quotient is 9.82 N/kg, which rounds to the 9.8 m/s² used throughout this course. The two units, N/kg and m/s², are the same thing: a newton is a kilogram-meter per second squared. The small mismatch with the more precise 9.81 comes from rounding Earth's mass and radius.

Why everything falls at the same rate. Compare g = F_g/m with Newton's second law, a = F_net/m. For an object in free fall with no air resistance, the only force is gravity, so a = F_g/m = g. The mass cancels: a heavier object feels a larger gravitational force, but it also needs a proportionally larger force to accelerate, and the two effects exactly balance. That is why a bowling ball and a marble dropped together hit the floor together in a vacuum, and why a feather lags only because of air.

Field strength elsewhere. At the altitude of a low orbit, about 400 km up, r = 6.77 × 10⁶ m and g = 3.98 × 10¹⁴ ÷ (6.77 × 10⁶)² = 8.7 N/kg, still 89 percent of the surface value. Orbiting astronauts float not because gravity is absent but because they and their spacecraft are falling together around Earth; the orbit is treated in the circular-motion lesson. At the Moon's surface, with the Moon's mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m, g = 6.674 × 10⁻¹¹ × 7.35 × 10²² ÷ (1.74 × 10⁶)² = 1.6 N/kg, about one sixth of Earth's; a 60 kg astronaut has the same mass there but weighs about 97 N instead of 588 N.

Net force between two bodies. A 1000 kg probe sits on the line between Earth and the Moon, 3.0 × 10⁸ m from Earth's center; Earth and Moon centers are about 3.84 × 10⁸ m apart, so the probe is 0.84 × 10⁸ m from the Moon's center. Force from Earth: 6.674 × 10⁻¹¹ × 5.97 × 10²⁴ × 1000 ÷ (3.0 × 10⁸)² = 3.98 × 10¹⁷ ÷ 9.0 × 10¹⁶ = 4.43 N toward Earth. Force from the Moon: 6.674 × 10⁻¹¹ × 7.35 × 10²² × 1000 ÷ (8.4 × 10⁷)² = 4.91 × 10¹⁵ ÷ 7.06 × 10¹⁵ = 0.70 N toward the Moon. The forces are opposite, so the net force is 4.43 − 0.70 = 3.7 N toward Earth. The forces balance where the distances satisfy M_E/r_E² = M_M/r_M², so r_E/r_M = √(M_E/M_M) = √81.2 = 9.0; the balance point is 9.0/10 of the way to the Moon, about 3.46 × 10⁸ m from Earth's center. Beyond that point the Moon's pull wins.

Scales and apparent weight. A bathroom scale does not measure gravitational force directly; it measures the normal force between the object and the surface supporting it. When nothing accelerates, the normal force equals the weight and the reading is correct. In an accelerating elevator the reading differs from the weight and is called the apparent weight; the previous lesson calculates it with free-body diagrams. Weight itself, mg, is unchanged.

Investigation: field strength as force per kilogram. Materials: a spring scale reading in newtons (0 to 5 N or 0 to 10 N), slotted masses of 0.100, 0.200, 0.300, 0.500 kg, and a stand. Steps: hang each mass from the scale at rest, read the force, and compute F/m for each. Record: a table of mass, force and F/m. Analyze: plot force against mass; the slope is g. A result that supports the claim: every F/m near 9.8 N/kg and a straight line through the origin, showing that the field strength does not depend on the mass placed in it. Then, in a gravity force simulation that lets you set two masses and their separation, record the force for one mass doubled, for the separation doubled and for the separation halved, and check that the force doubles, quarters and quadruples. Safety: use masses within the scale's range and keep feet clear. What the model cannot show: a spring scale in the classroom cannot detect the change of g with altitude (a few meters changes it by about one part in a million) or the pull of the masses on each other; the simulation shows the law but not the tiny real forces.

Worked example 1 — force between the Earth and the Moon. F = 6.674 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²² ÷ (3.84 × 10⁸)². Numerator: 6.674 × 5.97 × 7.35 × 10³⁵ = 2.93 × 10³⁷. Denominator: 1.47 × 10¹⁷. Force: 2.0 × 10²⁰ N, acting on each body toward the other. The Moon's acceleration from this force is F/M_M = 2.0 × 10²⁰ ÷ 7.35 × 10²² = 2.7 × 10⁻³ m/s², which is also g_Earth at the Moon's distance: 9.8 ÷ 60.3² = 2.7 × 10⁻³ m/s².

Worked example 2 — g at twice Earth's radius. At r = 2R, g = GM/(2R)² = (GM/R²)/4 = 9.8 ÷ 4 = 2.45 N/kg. A 50 kg object there weighs 50 × 2.45 = 123 N instead of 490 N. At 3R the field is 9.8 ÷ 9 = 1.09 N/kg, and at 4R it is 0.61 N/kg. The field never reaches zero; it only weakens.

Worked example 3 — the mass of a planet from g and radius. A probe measures g = 3.7 N/kg at the surface of a planet of radius 3.4 × 10⁶ m. From g = GM/r², M = gr²/G = 3.7 × (3.4 × 10⁶)² ÷ 6.674 × 10⁻¹¹ = 3.7 × 1.156 × 10¹³ ÷ 6.674 × 10⁻¹¹ = 6.4 × 10²³ kg. The method needs only a surface measurement of g and the radius, which is how planetary masses are estimated from landers and orbiters.

Worked example 4 — negligible or not? A 2000 kg truck parks 3.0 m from a 1000 kg car (center to center). F = 6.674 × 10⁻¹¹ × 2000 × 1000 ÷ 9.0 = 1.5 × 10⁻⁵ N. Compare with the static friction that holds the car's tires: even 1 N would be 70,000 times larger. The gravitational pull is real and calculable but negligible against every contact force present, which is why gravity between everyday objects is left out of free-body diagrams in ordinary mechanics problems. It is included only when it is the interaction being measured, as in a torsion-balance measurement of G.

Representation task: a field diagram with numbers. Draw Earth as a circle of radius 2 cm. At distances of 2, 3, 4 and 5 cm from the center along four directions (up, down, left, right), draw field arrows pointing toward the center with lengths proportional to 9.8, 4.4, 2.45 and 1.6 N/kg for 1R, 1.5R, 2R and 2.5R (use 1 cm for 9.8 N/kg). Label each arrow with its field strength and write the calculation for at least two of them from g = GM/r². Under the diagram, explain in two sentences why the arrows point inward and why they shorten by a factor of four when the distance doubles.

Independent performance: force between two massive bodies. For a 500 kg satellite placed 1.0 × 10⁸ m from Earth's center on the line toward the Moon (Earth-Moon distance 3.84 × 10⁸ m, so 2.84 × 10⁸ m from the Moon's center), calculate the force from each body, the net force and its direction, then find the point where the two forces balance. Success criteria: all quantities in SI units with powers of ten handled correctly; the two forces are shown as opposite and the net is a difference, not a sum; the answer for the balance point is checked by substituting back and showing the two forces agree to two significant figures; and a final sentence explains why a satellite there is not "beyond Earth's gravity".