2.6 Gravitational Force

Picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. A ball dropped from a roof

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You stand on the roof of a 45 m building and let go of a tennis ball. Nobody pushes it. Nothing touches it. Yet it speeds up, faster and faster, and hits the ground about 3 seconds later.

Later that year an astronaut steps onto the Moon. Her body is the same body. Her mass is the same 80 kg. But a bathroom scale on the Moon would read only about one sixth of what it reads at home.

What pulls the ball down, how strong is that pull, and why is it weaker on the Moon?

2. The same astronaut on Earth and on the Moon

3. Gravity reaches across space

Quick check. An astronaut has a mass of 70 kg on Earth. What is her mass on the Moon?

4. A heavy rock and a light ball

5. Why don't two friends pull together?

6. Distance counts from the center

Quick check. Gravity pulls a probe with 900 N on a planet's surface. The probe flies out until it is three times as far from the planet's center. How hard does gravity pull now?

7. Check yourself

Think of your answer first, then tap to see it.

a) With no air, does a 2 kg rock fall faster than a 1 kg rock?

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No. Gravity pulls the 2 kg rock twice as hard, but it has twice the mass. Both speed up at 9.8 m/s².

b) Two asteroids move until they are three times as far apart. What happens to the gravity between them?

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It drops to one ninth, because F depends on 1 ÷ r² and 3² = 9.

c) What is the weight of a 50 kg student on Earth?

Show answer

W = m g = 50 kg × 9.8 m/s² = 490 N, pointing down.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

1. With no air, you drop a 5 kg ball and a 1 kg ball at the same moment from the same height. What happens?

2. Two asteroids pull on each other by gravity. They drift until they are twice as far apart. The pull is now:

3. What is the weight of a 10 kg box on Earth?

The idea in plain words

Gravity is a pull between any two objects that have mass. It acts at a distance: nothing has to touch. The pull is tiny for everyday objects. It only becomes big when one object is huge, like a planet.

Weight is the name for the gravitational force a planet exerts on you. Near Earth's surface it is

Fg = m g    with g = 9.8 N/kg = 9.8 m/s² near Earth

Mass (kg) is how much stuff you are made of. It is the same everywhere. Weight (N) is a force. It changes from planet to planet because g changes.

Trap: saying an astronaut has less mass on the Moon, or giving a weight in kilograms. Instead: mass (kg) is the same everywhere; weight (N) = mg changes with g: 80 kg is 784 N on Earth and 130 N on the Moon.

The gravitational field g tells you the force on each kilogram at a place: g = Fg / m. Its unit N/kg is the same as m/s². That is why every object in free fall (no air resistance) has the same acceleration g: a heavier ball gets a bigger pull, but it also has more mass to speed up, and the two effects cancel.

Newton's law of gravitation

For any two objects with masses m1 and m2, whose centers are a distance r apart:

Fg = G m1 m2 / r²    G = 6.67 × 10−11 N·m²/kg²

Trap: forgetting to square r, so doubling the distance only halves the force. Instead: F depends on 1/r²: double r and the force drops to one quarter; triple r and it drops to one ninth.

Put a small mass m on the surface of a planet with mass M and radius R. The force is G M m / R², and it is also m g. So the field at the surface is

g = G M / R²

m₁ m₂ F on m₁ F on m₂ r (center to center)

g on other worlds (surface values): Moon 1.62 m/s², Mars 3.71 m/s², Earth 9.8 m/s², Jupiter about 24.8 m/s².

Worked with numbers: the ball, the astronaut and the space station

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  1. Weight of a 0.15 kg ball on Earth: Fg = m g = 0.15 kg × 9.8 m/s² = 1.5 N down. Near the surface, weight is just mass times the field g.
  2. Time for the ball to fall 45 m (up positive, starts at rest): −45 = −½ (9.8) t², so t² = 90 / 9.8 = 9.18 s² and t = 3.0 s. The only force is weight, so a = Fg/m = g, a constant acceleration. Use Unit 1's Δy = v₀t + ½at².
  3. Speed at the ground: v = g t = 9.8 × 3.03 = 29.7 m/s (check: v² = 2 g h = 2 × 9.8 × 45 = 882, v = 29.7 m/s). Two different formulas give the same answer, a good sign the numbers are right.
  4. Astronaut, 80 kg. On Earth: 80 × 9.8 = 784 N. On the Moon: 80 × 1.62 = 130 N. Mass in both places: 80 kg. Mass does not change. Only g changes, so only weight changes. 784 / 130 ≈ 6.
  5. Check Earth's g from the law of gravitation: g = G M / R² = (6.67×10−11)(5.97×1024 kg) / (6.37×106 m)² = 3.98×1014 / 4.06×1013 = 9.8 m/s². The "9.8" is not magic: it comes from Earth's mass and radius.
  6. On the space station, 400 km up: r = 6.37×106 + 0.40×106 = 6.77×106 m, so g = 3.98×1014 / (6.77×106)² = 8.7 m/s². The astronaut's weight there is 80 × 8.7 = about 700 N. Gravity is still about 89% as strong as on the ground. Astronauts float because they and the station fall together (see 2.9), not because gravity is gone.

Try it: animations

Up is positive in every animation. Predict first, then press Play. Watch the free body diagram next to each scene.

1. Ball dropped from a building

Change the height, the mass and the field g. Does a heavier ball land sooner? Try g = 1.62 (Moon) and 24.8 (Jupiter).

Free body diagram of the ball

Only one force acts once the ball is let go: its weight m g, straight down. The arrow grows with mass, but the acceleration stays g.

2. An astronaut jumps on the Moon

Same jump speed, different worlds. Set g to 9.8 (Earth), 3.71 (Mars) or 1.62 (Moon).

Free body diagram in the air

Weight = m g. On the Moon the arrow is about 1/6 as long, so the astronaut rises about 6 times higher and stays up about 6 times longer.

3. Newton's law of gravitation: carry a probe away from a planet

A (very fast, made-up) rocket carries a probe straight away from the planet at a steady 1000 km/s. Watch the gravitational force on the probe as the distance grows. Distances are measured from the planet's center, in Earth radii (1 Earth radius = 6370 km).

Free body diagram of the probe (gravity only)

At 2 radii the force is 1/4 of the surface value; at 3 radii, 1/9; at 10 radii, 1/100. It gets small fast but never reaches zero.

Examples

Example 1: mass and weight basic

A backpack has a mass of 6.0 kg. (a) What is its weight on Earth? (b) What are its mass and weight on Mars, where g = 3.71 m/s²?

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  1. (a) Fg = m g = 6.0 × 9.8 = 59 N. Weight is a force, so the unit is newtons.
  2. (b) Mass is still 6.0 kg. Weight = 6.0 × 3.71 = 22 N. Mass belongs to the object; weight depends on where the object is.

Example 2: twice as far medium

A satellite weighs 2000 N on the launch pad. What is the gravitational force on it at a height above the surface equal to one Earth radius?

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  1. Distance from the center goes from R to R + R = 2R. r is center to center, not height above the ground.
  2. F is proportional to 1/r², so Fnew = 2000 N × (R/2R)² = 2000 / 4 = 500 N. Doubling r divides the force by 2² = 4. No need for G or M: use a ratio.

Example 3: a new planet AP

Planet X has twice the mass of Earth and twice the radius. (a) Find g on its surface. (b) A rock is dropped from 10 m. How long does the fall take on Planet X and on Earth?

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  1. g = G M / R². With 2M and 2R: gX = G(2M)/(2R)² = (2/4) G M/R² = ½ × 9.8 = 4.9 m/s². Mass doubling helps by 2; radius doubling hurts by 4.
  2. Planet X: 10 = ½ (4.9) t², t² = 4.08 s², t = 2.0 s. Dropped means v₀ = 0.
  3. Earth: 10 = ½ (9.8) t², t² = 2.04 s², t = 1.4 s. Half the g gives √2 times the time: 1.43 × 1.41 = 2.02 s. Consistent.

Example 4: why don't parked cars pull together? AP

Two 1000 kg cars are parked with their centers 2.0 m apart. Find the gravitational force between them and compare it with the weight of one car.

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  1. F = G m₁ m₂ / r² = (6.67×10−11)(1000)(1000)/(2.0)² = 6.67×10−5 / 4 = 1.7×10−5 N. Plug in with SI units.
  2. Weight of one car = 1000 × 9.8 = 9800 N. Ratio: 1.7×10−5 / 9800 ≈ 2×10−9. Gravity between everyday objects is real but far too small to beat friction. Only planet-sized masses make big gravitational forces.

Practice (AP style)

1. An astronaut has a mass of 70 kg on Earth. On the Moon (g = 1.62 m/s²) her mass is

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70 kg. Mass does not depend on location. Her weight changes (686 N on Earth, 113 N on the Moon), but 113 and 686 are newtons, not kilograms. 11 kg comes from wrongly turning the Moon weight back into kilograms with Earth's g: 113 N / 9.8 ≈ 11.6 kg.

2. Two small asteroids attract each other with force F. The distance between their centers is tripled. The new force is

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F/9. F ∝ 1/r². Tripling r multiplies the force by 1/3² = 1/9. F/3 forgets to square.

3. Both masses are doubled and the distance between them is also doubled. The gravitational force

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Stays the same. F ∝ m₁m₂/r² → (2)(2)/(2)² = 4/4 = 1.

4. A 1 kg ball and a 5 kg ball are dropped together from the same height. Air resistance is negligible. Which statement is correct?

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a = Fg/m = m g / m = g for both. The forces are NOT equal (9.8 N vs 49 N), which is why the last choice is wrong even though its conclusion is right.

5. The Earth exerts a 2.0×1020 N force on the Moon. The force the Moon exerts on the Earth is

Show answer

Newton's third law: the forces are equal in size and opposite in direction. The Earth is pulled toward the Moon. The Earth accelerates much less only because its mass is much larger.

6. At what height above Earth's surface (radius R) is g equal to 9.8/4 = 2.45 m/s²?

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g ∝ 1/r². One quarter of the surface value needs r = 2R from the center, which is a height of R above the surface. "2R" is the distance from the center, a common trap.

7. Which graph shape best shows the gravitational force on a probe versus its distance r from a planet's center (for r larger than the radius)?

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F ∝ 1/r²: the drop is fast at first and slows down; F gets close to zero but never reaches it. A graph of F versus 1/r² would be a straight line through the origin.

8. (Short free response: mathematical routines) On a newly found planet, a student drops a ball from 20.0 m and measures a fall time of 2.60 s. The planet's radius is 4.0×106 m. (a) Find g on the planet. (b) Find the planet's mass. (c) Would a 2 kg ball take more, less or the same time? Explain in one or two sentences.

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  1. (a) Δy = ½ g t² → g = 2Δy/t² = 2(20.0)/(2.60)² = 40.0/6.76 = 5.9 m/s².
  2. (b) g = G M/R² → M = g R²/G = (5.92)(4.0×106)²/(6.67×10−11) = (5.92)(1.6×1013)/(6.67×10−11) = 1.4×1024 kg.
  3. (c) The same time. The gravitational force is proportional to the mass, so a = F/m = g for any mass (air resistance negligible).

Scoring idea: 1 point for the correct kinematics equation with v₀ = 0, 1 for g, 1 for using g = GM/R², 1 for M with units, 1 for "same time" with the reason that F and m both scale.

9. (Short free response) A student says: "Astronauts on the space station float because there is no gravity in space." Using numbers, explain what is wrong. The station orbits 400 km above Earth's surface (Earth radius 6370 km).

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g at the station = 9.8 × (6370/6770)² = 9.8 × 0.885 = 8.7 m/s², about 89% of the surface value. Gravity is very much there. Astronauts feel weightless because they and the station are both falling around the Earth with the same acceleration, so the floor does not push on them (no normal force). That is free fall, not zero gravity.

10. (Short free response: translation between representations) A probe of mass m is at distance r from the center of a planet. At r = R (the surface) the gravitational force on it is 900 N. (a) Copy and complete the table: r = R, 2R, 3R; F = 900 N, ?, ?. (b) Which description matches the graph of F against r? (A) a straight line sloping down; (B) a curve that drops steeply, then flattens out; (C) a horizontal line; (D) a straight line sloping up. (c) What should be plotted on the horizontal axis to make the graph a straight line through the origin, and what does its slope equal?

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  1. (a) F is proportional to 1/r²: at 2R, F = 900 / 4 = 225 N; at 3R, F = 900 / 9 = 100 N.
  2. (b) (B). Doubling r cuts F to one quarter, tripling it to one ninth: big drops at first, then smaller and smaller ones. A straight line would be wrong because F does not fall by the same amount for each step in r.
  3. (c) Plot F against 1/r². Since F = GMm (1/r²), the slope is GMm.

Point guide (4 points):

  • 1 point: 225 N and 100 N.
  • 1 point: picks (B).
  • 1 point: explains the shape with the inverse square.
  • 1 point: F vs 1/r² with slope GMm.

11. (Short free response: experimental design and analysis) A student wants to show that weight is proportional to mass and to measure g. She has a set of masses and a force sensor (a spring scale) that hangs from a stand. (a) Describe her procedure. (b) Her data: 0.10 kg: 0.98 N; 0.20 kg: 1.95 N; 0.30 kg: 2.95 N; 0.40 kg: 3.92 N; 0.50 kg: 4.90 N. What should she plot, and what does the slope tell her? Calculate it. (c) If she repeated the experiment on the Moon (g = 1.62 m/s²), how would the graph change? Would a pan balance that compares masses give different readings there?

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  1. (a) Hang each mass from the sensor, wait until it is still (a = 0, so the sensor reading equals the weight), and record the reading. Use at least five masses and repeat each.
  2. (b) Plot weight F (vertical) against mass m (horizontal). A straight line through the origin shows F is proportional to m. Slope = (4.90 − 0.98) / (0.50 − 0.10) = 3.92 / 0.40 = 9.8 N/kg = g.
  3. (c) Still a straight line through the origin, but the slope would be 1.62 N/kg: each weight about one sixth as big. A pan balance would read the same masses, because it compares two masses that both feel the same smaller g. Mass does not change; weight does.

Point guide (4 points):

  • 1 point: records force for several masses with the object at rest.
  • 1 point: plots F vs m and says a straight line through the origin shows proportionality.
  • 1 point: slope 9.8 N/kg identified as g.
  • 1 point: Moon slope 1.62 N/kg, balance readings unchanged (mass vs weight).

12. (Short free response: qualitative/quantitative translation) Planet X has twice the mass of Earth and twice the radius of Earth. (a) Derive an expression for the gravitational field g at the surface of a planet in terms of G, the planet's mass M and its radius R. (b) Without numbers, predict whether g on Planet X is larger, smaller or the same as on Earth, and by what factor. (c) Calculate g on Planet X (use 9.8 m/s² for Earth) and the weight there of a 60 kg person. (d) Is your answer to (c) consistent with (b)?

Show answer
  1. (a) mg = GMm / R², so g = GM / R².
  2. (b) Doubling M doubles g; doubling R divides g by 4 (R is squared). Net factor 2 ÷ 4 = ½: g on Planet X is half of Earth's.
  3. (c) gX = 9.8 × 2 ÷ 4 = 4.9 m/s². Weight = 60 × 4.9 = 294 N.
  4. (d) Yes: 4.9 is half of 9.8, the factor of ½ predicted in (b).

Point guide (4 points):

  • 1 point: g = GM / R².
  • 1 point: factor ½ with the R² reason.
  • 1 point: 4.9 m/s² and 294 N.
  • 1 point: links (c) back to (b).

Common mistakes

Mixing up mass and weight

The mistake: "My weight is 60 kg" or "on the Moon my mass is smaller".

Why it is wrong: mass is in kg and never changes; weight is a force in N and equals m g.

How to spot it: check the unit. If a "weight" is in kg, or a "mass" changes between planets, fix it.

Measuring r from the surface

The mistake: using the height above the ground as r in G M m / r².

Why it is wrong: the law uses the distance between centers. Near Earth, r = R + h.

How to spot it: if you get a huge force for a satellite (because r is small), you probably forgot to add R.

Forgetting to square r

The mistake: "Twice as far means half the force."

Why it is wrong: gravity follows an inverse square law. Twice as far gives one quarter.

How to spot it: in ratio problems, always write (rold/rnew)² and check the exponent.

"Heavier things fall faster"

The mistake: thinking the bigger gravitational force gives a bigger acceleration.

Why it is wrong: a = F/m = mg/m = g. More force, but also more mass to accelerate.

How to spot it: in any free-fall problem without air resistance, the mass should cancel. If it does not, recheck.

"There is no gravity in space"

The mistake: believing that orbiting astronauts float because gravity is zero.

Why it is wrong: at 400 km, g is still 8.7 m/s². Gravity keeps the station in orbit. Floating is free fall.

How to spot it: if gravity were zero, nothing would curve the station's path, and it would fly off in a straight line.

Sign of g

The mistake: writing g = −9.8 m/s² and then a = −g, getting +9.8.

Why it is wrong: g is a size (9.8 N/kg). The direction goes into the sign of a: with up positive, a = −g = −9.8 m/s².

How to spot it: a dropped ball with up positive must have negative velocity and negative acceleration.