Gravitational Force
1. Every mass pulls every other mass
Gravity is the one force in this course that does not need contact. Two objects with mass attract each other along the line joining their centers of mass, always attract, never repel, and the size of the pull is
Fg = G m1m2 / r²G = 6.67 × 10⁻¹¹ N·m²/kg²
The pair is a third-law pair, so the Earth pulls on you exactly as hard as you pull on the Earth. You accelerate noticeably and the Earth does not, for the reason Topic 2.3 gave.
The r is measured between centers of mass, not between surfaces, and that single detail settles most of the questions people find confusing. Standing on a mountain barely changes your weight because the mountain adds two kilometers to six thousand four hundred.
Slide the object away from the Earth and watch both the arrow and the curve. The distance is counted in Earth radii from the center, so the leftmost position is standing on the ground. The fall is much faster than most people expect.
2. The field, and why it is a useful idea
Rather than recompute the force for every object you might put at a place, compute the pull per kilogram once and keep it. That is the gravitational field.
g = Fg / m = G M / r²in newtons per kilogram
Near the surface of the Earth it is about 10 N/kg, and if gravity is the only force on an object, the acceleration in meters per second squared is numerically the same number. That is not a coincidence, it is the second law: divide the force by the mass and the mass cancels, which is why a feather and a hammer fall together in a vacuum.
The weight of an object is the gravitational force an astronomical body exerts on it, Fg = mg, measured in newtons. It is not the mass, and it is not what a bathroom scale reads.
3. When you can treat gravity as constant
The inverse square law is always true, and almost never worth using near the ground. Going from sea level to the top of a very tall building changes the field by a few hundredths of a percent, so for anything happening within a few kilometers of the surface, treating g as constant is not an approximation you should apologize for.
Reach for the full expression when the distance from the center of the Earth changes by a noticeable fraction: satellites, orbits, and anything phrased in terms of Earth radii.
4. Apparent weight is the normal force
Stand on a scale. The scale does not measure the Earth pulling on you. It measures how hard it has to push up on you, which is the normal force, and it reports that number. Your apparent weight is that push.
When nothing accelerates, the two arrows balance and the scale happens to read your weight, which is why the distinction is easy to miss. Accelerate and they come apart at once.
A person on a scale in an elevator. Change the acceleration and watch the two readouts: one of them never moves. Push the slider all the way down, to an acceleration of 9.8 meters per second squared downward, and see what the scale says.
A system appears weightless when gravity is the only force on it, which is exactly the situation of an astronaut in orbit. There is nothing weak about the gravity there: at the height of the space station the field is roughly nine tenths of its surface value. The astronaut floats because the station and everything in it are falling together, so no floor is pushing on anybody.
5. What a sphere of mass actually does
Everything above has treated a planet as though its mass sat at a point. That is not obvious. A planet is a huge collection of differential masses at different distances and different angles, and the honest calculation adds up the pull from every one of them. Newton’s shell theorem is the result of doing that for a uniform spherical shell, and it says two remarkable things.
Outside the shell, the pull is exactly what you would get if the entire mass of the shell sat at its center. Not approximately. Exactly, at any distance outside it, however close to the surface.
Inside the shell, the net pull is exactly zero, everywhere, not just at the middle. The nearby part of the shell pulls harder but there is less of it; the far part is weaker but there is more. The two effects cancel perfectly.
Stack shells to build a solid sphere and both results carry over. From outside, a uniform sphere acts as a point at its center. From inside, every shell above you contributes nothing, so only the mass closer to the center than you are pulls at all:
mpartial = ρ · (4/3)π r³with only that mass counted
Put that into the inverse square law and something unexpected happens. The enclosed mass grows as r³ and the law divides by r², so what survives is proportional to r:
Fg = −k rinside a uniform sphere
That is Hooke’s law with a different name on it. Drop a ball down a tunnel bored through a uniform planet and it oscillates about the center, exactly as a mass on a spring does. This course does not ask you to derive the shell theorem, but it does expect you to use it.
Move the test mass from the center of a uniform planet out into space. Inside, the shaded region is the only mass that pulls on it. Watch the graph: a straight line out to the surface and an inverse square beyond, meeting exactly at the surface.
6. Two kinds of mass that turn out to be one
Inertial mass is the mass in the second law, the property that decides how little an object accelerates for a given push. Gravitational mass is the mass in the law of gravitation, the property that decides how strongly it is attracted. There is no obvious reason for these to be the same number, any more than electric charge has to equal mass.
They have been measured against each other to extraordinary precision and no difference has ever been found. That equivalence is why all objects fall with the same acceleration, and it is the starting point of general relativity.
7. Reading it wrong
Measuring r from the surface. It is from the center. A satellite 6400 km up is at two Earth radii, not one.
Calling the scale reading your weight. It is the normal force. In an accelerating elevator the two differ, and the whole point of the topic is that they are separate quantities.
Saying there is no gravity in orbit. There is almost as much as here. Orbiting is falling, continuously, and missing.
Using the full inverse square law inside a planet. It gives an infinite force at the center, which should be enough of a warning. Inside a uniform sphere the force is linear in the distance, not inverse square, because most of the mass has stopped counting.
Check yourself
1. A satellite orbits at a distance of 2 Earth radii from the center. Compare the gravitational field there with the field at the surface, and give the acceleration of the satellite.
Twice the distance means a quarter of the field, so about 2.5 N/kg. Gravity is the only force acting, so the acceleration is numerically the same, 2.5 m/s², directed toward the center of the Earth. That acceleration is what keeps it curving around rather than flying off.
2. A 60 kg person stands on a scale in an elevator accelerating upward at 2 m/s². Find the scale reading, and state what the person’s weight is.
The scale reads the normal force: Fn = m(g + a) = 60(9.8 + 2) = 708 N. The weight is unchanged at 60 × 9.8 = 588 N, because the Earth has not changed and neither has the person. Two different numbers for two different forces.
3. An astronaut on the space station appears weightless. A student concludes that there is no gravity at that altitude. Correct them with a number and a reason.
The station orbits at roughly 1.05 Earth radii, so the field is about 9.8/1.05², close to 8.9 N/kg, nearly what it is on the ground. The astronaut floats because the station, the floor and the astronaut are all accelerating downward together, so no normal force is needed. Apparent weight is zero; weight is not.
4. Explain why a hammer and a feather dropped together on the Moon land together, using both laws that the argument needs.
The gravitational force on each is proportional to its own mass, F = mg. The acceleration is that force divided by the same mass, so the mass cancels and every object gets the same acceleration. It needs the law of gravitation to supply a force proportional to mass and the second law to divide by mass, and it works on the Moon rather than on Earth only because there is no air there to spoil it.
5. A tunnel is bored straight through the center of a uniform planet of radius R. Describe the gravitational force on a ball at distance R/2 from the center and at the center itself, and say what kind of motion the ball performs if it is released at the surface.
At R/2 only the mass within that radius pulls, which is (1/2)³ = 1/8 of the planet, and the force is half its surface value: 1/8 of the mass divided by (1/2)² of the distance squared. At the center the enclosed mass is zero and so is the force. The force is proportional to the distance and directed toward the center, which is Hooke’s law, so the ball oscillates back and forth through the planet in simple harmonic motion.
6. An elevator cable snaps and the car falls freely. Describe the reading on a scale inside, the weight of the person on it, and the free-body diagram, and say which of the three changed.
The scale reads zero. The weight is unchanged. The diagram loses the normal arrow entirely and keeps the weight arrow, so it goes from two forces to one. Only the normal force changed, and it changed because the floor no longer has to support anybody: everything is accelerating downward at the same rate.
Topic 2.7, Kinetic and Static Friction. The normal force has just done a lot of work in this topic. The next one gives it a second job, because friction is proportional to it, and that is where the surfaces start to matter.