Mass vs Weight: What Is the Difference?

Mass is a property of the object itself, measured in kilograms, and it is the same everywhere. Weight is the gravitational force a planet exerts on the object, measured in newtons, and it changes when the object moves somewhere the gravitational field is different. Weight equals mass times g.

AP Physics: Unit 2 (topics 2.6 Gravitational Force). This comparison sits inside AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over an estimated 22 to 27 class periods. Topic 2.6 Gravitational Force holds both halves of the distinction. EK 2.6.A.3 defines weight as the gravitational force exerted by an astronomical body on a relatively small nearby object and gives the derived equation Weight = F_g = mg. Learning objective 2.6.D, describe inertial and gravitational mass, is supported by EK 2.6.D.1 (inertial mass, or inertia, is a property that determines how much an object's motion resists changes when interacting with another object), EK 2.6.D.2 (gravitational mass is related to the force of attraction between two systems with mass) and EK 2.6.D.3 (the two have been experimentally verified to be equivalent). EK 2.6.A.2.ii states that when gravity is the only force on an object, its acceleration in m/s^2 is numerically equal to the gravitational field strength in N/kg. EK 2.6.C.1 through 2.6.C.4 cover apparent weight and the equivalence principle. EK 2.6.B.1 sets the condition under which the gravitational force may be treated as constant. The CED lists no boundary statement under Topic 2.6; the g value used on the exam comes from a Topic 1.3 boundary statement. Suggested skills for Topic 2.6 are 1.A, 2.A, 2.D and 3.C.

The distinction, stated once

Mass belongs to the object. Weight is something a planet does to it. Change nothing about the object and carry it somewhere else, and the mass reads the same number while the weight reads a different one.

The AP Physics 1 CED defines the two in separate places inside the same topic, Topic 2.6 Gravitational Force, and the definitions are built differently.

Weight names an interaction. EK 2.6.A.3 says the gravitational force exerted by an astronomical body on a relatively small nearby object is called weight, and gives the derived equation $\text{Weight} = F_g = mg$ . Read what has to be present for that sentence to mean anything: an astronomical body, an object near it, and a force between them. Weight is not a possession. It is the name of a pull, and it takes two participants.

Mass names a property. EK 2.6.D.1 says objects have inertial mass, or inertia, a property that determines how much an object's motion resists changes when interacting with another object. EK 2.6.D.2 says gravitational mass is related to the force of attraction between two systems with mass. One object, one number, no second participant required.

So the compressed version is four contrasts. Kilograms against newtons. Scalar against vector. A property against a force. The same on every planet against different on every planet.

This page is about the difference and what it costs when you miss it. If you want each side on its own, the mass and weight glossary entries define them one at a time.

Mass vs weight, side by side

	Mass	Weight
What it is	A property of the object: how much it resists acceleration, and how strongly it participates in gravity	The gravitational force an astronomical body exerts on the object
CED essential knowledge	2.6.D.1, 2.6.D.2	2.6.A.3
Symbol	$m$ , or $M$ for the larger of two bodies	$F_g$
SI unit	kilogram, kg	newton, N
Scalar or vector	Scalar	Vector
Direction	It has none	Toward the center of mass of the attracting body
Changes if you move the object	No	Yes, wherever the gravitational field strength differs
Value far from every planet	Unchanged	Falls off as $1/r^2$ , so it shrinks without ever reaching zero
How you measure it	A balance, or apply a known force and measure the acceleration	A spring scale or force sensor, when nothing is accelerating
Role in Newton's second law	It is the $m$ in $\vec{a}_{\text{sys}} = \vec{F}_{\text{net}} / m_{\text{sys}}$	It is one of the forces you add up to get $\vec{F}_{\text{net}}$
Near-surface equation	No defining equation, it is a measured input	$F_g = mg$
General equation	None	$\lvert \vec{F}_g \rvert = \frac{G m_1 m_2}{r^2}$
On the AP Physics 1 equation sheet	kilogram appears in the unit-symbols table, and the symbol key reads $m$ = mass	$\lvert \vec{F}_g \rvert = \frac{G m_1 m_2}{r^2}$ is printed, $F_g = mg$ is not
The everyday sentence that gets it wrong	Nobody says "my mass is 70 newtons"	"My weight is 70 kilograms" states a mass

The last row is the whole reason this confusion exists. Everyday English uses the word weight for the number a bathroom scale shows, that number is printed in kilograms, and kilograms are a unit of mass. So the word and the unit come from opposite sides of the table, and both of them get taught before anyone mentions it.

The case that separates them: keep the object, change the place

One 6.0 kg toolbox, four locations, nothing done to the toolbox. Only the gravitational field strength $g$ changes from row to row, and the CED prints a value only for Earth's surface, so the others are supplied here as problem data.

Where the toolbox is	Field strength	Mass	Weight $F_g = mg$	What a spring scale under it reads
On Earth's surface	9.8 N/kg	6.0 kg	58.8 N	58.8 N
On a world where the field is weaker	1.6 N/kg	6.0 kg	9.6 N	9.6 N
Inside an orbiting spacecraft	8.7 N/kg	6.0 kg	52.2 N	0 N
Far from every star and planet	about 0	6.0 kg	about 0 N	0 N

The mass column never moves. That is what it means for mass to be a property of the object.

The weight column tracks the field, and row two is the one people expect: a weaker field, a smaller pull, and for this object a force smaller by the factor $9.8 / 1.6 = 6.125$ . Multiply the mass by the local field strength and you are done.

Row three is the one that catches people. The field inside a low orbit is not much weaker than at the surface, so the gravitational force on the toolbox there really is about 52 N. The scale still reads zero, because the scale is falling alongside the toolbox and has nothing to push with. That zero is not the weight. It is the apparent weight, which EK 2.6.C.1 identifies with the magnitude of the normal force, and EK 2.6.C.3 says a system appears weightless when there are no forces exerted on the system or when the force of gravity is the only force exerted on the system. The word is appears. The 52 N is still acting. What went to zero is the contact force, and weight vs normal force is where that split is worked out in full.

Row four is worth one sentence. Since $\lvert \vec{F}_g \rvert = G m_1 m_2 / r^2$ never actually reaches zero for finite $r$ , an object far from everything has a tiny weight rather than no weight. Its mass is still 6.0 kg.

Mass does two jobs, and nothing forces them to agree

This is the part of the distinction that is genuinely deep rather than a units problem, and the CED gives it its own learning objective, 2.6.D, describe inertial and gravitational mass.

Inertial mass is the $m$ in $\vec{a}_{\text{sys}} = \vec{F}_{\text{net}} / m_{\text{sys}}$ . It answers how hard this object is to speed up or slow down. EK 2.6.D.1 defines it as a property that determines how much an object's motion resists changes when interacting with another object. You can measure it with no gravity anywhere in the experiment: push, measure the acceleration, divide.

Gravitational mass is the $m$ in $F_g = mg$ and the $m_1$ and $m_2$ in $G m_1 m_2 / r^2$ . It answers how strongly this object joins in gravitational interactions. EK 2.6.D.2 states it in those terms: gravitational mass is related to the force of attraction between two systems with mass.

Those are two different questions, and there is no logical reason the answers should be the same number. Electric charge makes the point by contrast: an object's charge sets how strongly it takes part in electric forces and tells you nothing at all about how hard it is to accelerate. Gravity could have worked the same way, with a separate gravitational charge unrelated to inertia. It does not, and EK 2.6.D.3 records why we say so: inertial mass and gravitational mass have been experimentally verified to be equivalent.

That one experimental result is doing more work than it looks. Write the second law for an object with only gravity acting on it, keeping the two masses apart:

m_{\text{grav}}\, g = m_{\text{inert}}\, a

If the two are the same number they cancel and $a = g$ , the same for every object at that location whatever it is made of and however heavy it is. The mass cancellation you use in every free-fall problem is a measured fact about the universe, not a step of algebra. A brick and a pebble land together because of EK 2.6.D.3.

EK 2.6.A.2.ii says the same thing with the units made explicit: if the gravitational force is the only force exerted on an object, the observed acceleration of the object in $\text{m/s}^2$ is numerically equal to the magnitude of the gravitational field strength in N/kg at that location. That is also why the equation sheet prints $g$ twice, once as $9.8\ \text{m/s}^2$ and once as $9.8\ \text{N/kg}$ : two different quantities, one number, and the equality is the content.

The CED takes it one step further at EK 2.6.C.4, which states the equivalence principle: an observer in a noninertial reference frame is unable to distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field.

kg is not a unit of force, and N is not a unit of mass

The unit swap is how this confusion usually shows up on paper, so it is worth being blunt about what each unit is attached to.

A kilogram measures mass. It appears in the AP Physics 1 unit-symbols table, which lists eight units in all, with kilogram (kg) and newton (N) as separate entries. A newton measures force, and one newton is one kilogram metre per second squared, which is what $\vec{F}_{\text{net}} = m\vec{a}$ makes it.

So why does a bathroom scale read kilograms? Because it is doing a hidden division. The spring inside it responds to a force. The dial has been printed by taking that force and dividing by $9.8\ \text{N/kg}$ before you see it. That conversion is correct only where the field strength is $9.8\ \text{N/kg}$ and nothing is accelerating, and the scale has no way to check either condition. Worked example three below carries the same scale to a weaker field and watches it lie.

There is a unit that does measure force in kilograms, the kilogram-force, defined as the weight of one kilogram where $g$ is $9.8\ \text{N/kg}$ , so about 9.8 N. It is not part of SI and it does not appear on any of the four AP equation sheets. If you meet it outside physics class, read it as a force.

One small habit fixes most of the damage. Whenever you write a number, write its unit at the same time, and check that the unit matches the slot. The second law divides a force by a mass, so newtons over kilograms; $\Delta U_g = mg\Delta y$ multiplies kilograms by newtons per kilogram by metres, which leaves joules. The AP task verb Calculate is defined in the CED as performing mathematical steps to arrive at a final answer including algebraic expressions, properly substituted numbers, and correct labeling of units and significant figures, so the unit is part of the answer rather than decoration on it.

Where it costs a mark

Each of these is a scoring event rather than a philosophical point.

Reporting a weight in kilograms. Weight is a force and its unit is the newton. A 4.0 kg block has a weight of 39.2 N, and writing "the weight is 4.0 kg" answers a different question from the one asked.
Reporting a mass in newtons. The reverse error, and it usually arrives by substituting a weight into $\vec{a} = \vec{F}_{\text{net}}/m$ . If the number you divide by is in newtons, divide it by $g$ first.
Saying mass changes on the Moon. It does not. The gravitational field strength changes, so the weight changes with it.
Saying an astronaut in orbit has no weight. EK 2.6.C.3 says the system appears weightless when there are no forces exerted on the system or when the force of gravity is the only force exerted on the system. In orbit gravity is the only force, so the second clause applies, and gravity being the only force is very different from gravity being absent.
Using $F_g = mg$ where the field is not constant. EK 2.6.B.1 sets the condition: if the gravitational force between two systems' centers of mass has a negligible change as their relative position changes, the force can be treated as constant between the initial and final positions. A problem that moves an object a large fraction of a planetary radius fails that test, and you need $G m_1 m_2 / r^2$ instead.
Treating $F_g = mg$ as printed. It is not on the AP Physics 1 equation sheet. The CED lists it under EK 2.6.A.3 as a derived equation, so you write it out yourself.
Calling weight a scalar because $mg$ looks like plain multiplication. The sheet writes the gravitational force with vector notation and magnitude bars, $\lvert \vec{F}_g \rvert$ , precisely because the underlying quantity has a direction. Straight down, near a surface.
Letting mass go missing from a rotational or momentum expression because you substituted a weight. $\vec{p} = m\vec{v}$ and $K = \frac{1}{2}mv^2$ both take kilograms. Feed either of them 39.2 N and the units of the answer stop being kilogram metres per second or joules, which is the fastest way to catch yourself.
Assuming the exam uses 9.8. A Topic 1.3 boundary statement says that for all situations in which a numerical quantity is required for $g$ , the value $g \approx 10\ \text{m/s}^2$ will be used, then adds that students will not be penalized for correctly using the more precise commonly accepted values of $g = 9.81\ \text{m/s}^2$ or $g = 9.8\ \text{m/s}^2$ . The Table of Information prints 9.8. Every number on this page uses 9.8, and is g 9.8 or 10 covers what the choice does to your arithmetic.

When they seem to coincide, and why that lulls you

Near Earth's surface the two quantities are locked together by a single number. Multiply any mass in kilograms by $9.8\ \text{N/kg}$ and you have its weight in newtons; divide any weight by 9.8 and you have its mass. Twice the mass really does mean twice the weight, and the ordering of a list of objects by mass is the same as the ordering by weight.

That is why the distinction can go unnoticed for a long time. Every object in an introductory problem set sits in the same field, so the conversion factor is a constant you never have to think about, and the two quantities behave like the same thing measured in two currencies.

Three things break the illusion, roughly in the order they appear:

Leaving the surface. Another planet, an orbit, or a distance comparable to a planetary radius, and $g$ is no longer 9.8.
Acceleration. The gravitational force does not change, but the supporting force does, so a scale stops agreeing with the weight. That is the apparent-weight case, and EK 2.6.C.2 states it directly: if the system is accelerating, the apparent weight of the system is not equal to the magnitude of the gravitational force exerted on it.
An algebra step that needs one and not the other. Anything of the form $\vec{F}_{\text{net}}/m$ needs the mass, and anything that lists the forces on an object needs the weight. Substituting the wrong one is a units error that survives the whole page if you do not label as you go.

The reflex worth building is small. Write $mg$ rather than "weight", and keep the mass and the $g$ visible. Then when $g$ changes you can see where it enters, and when a problem hands you a force in newtons you can see that you have to divide before it can serve as a mass.

What the equation sheet gives you, and what it does not

The AP Physics 1 equation sheet handles mass and weight quite differently, and knowing which lines exist saves you from hunting for one that does not.

For mass, there is nothing to print. Mass is a measured input to the equations rather than a quantity you compute from other things, so it appears as a symbol inside other lines. The sheet's symbol key reads $m$ = mass, and separately $W$ = work. That second entry is worth registering, because no entry anywhere on the sheet assigns the letter $W$ to weight, so a $W$ you meet in a printed equation is a work.

For weight, the sheet prints the general two-body law and not the near-surface shortcut:

\lvert \vec{F}_g \rvert = \frac{G m_1 m_2}{r^2}

$F_g = mg$ does not appear anywhere on it. The CED lists that form under EK 2.6.A.3 as a derived equation, which means you write it out when you need it. The same is true of the field-strength relation $\lvert \vec{g} \rvert = \lvert \vec{F}_g \rvert / m = G M / r^2$ , listed as derived under EK 2.6.A.2.i and absent from the sheet.

The sheet does hand you $mg$ inside a different line, though. It prints $\Delta U_g = mg\Delta y$ for gravitational potential energy near a surface, and $U_G = -G m_1 m_2 / r$ for the general case, so the product of a mass and a field strength is sitting there in the energy section even though the weight itself is not in the force section.

In the constants box the sheet prints $g$ twice, on two separate lines: the magnitude of the acceleration due to gravity at Earth's surface, $g = 9.8\ \text{m/s}^2$ , and the magnitude of the gravitational field strength at Earth's surface, $g = 9.8\ \text{N/kg}$ . Those are the two faces of EK 2.6.A.2.ii, and the N/kg version is the one you want when you are computing a weight, because it multiplies a mass in kilograms and returns newtons without any unit gymnastics.

The full AP Physics 1 sheet is here. For the procedures that use these quantities rather than the distinction between them, how to find net force covers adding the weight into a force sum, how to draw a free-body diagram covers drawing it, and Topic 2.6 carries the CED framing for the whole topic.

One toolbox, three field strengths

A 6.0 kg toolbox is taken from Earth's surface, where the gravitational field strength is $9.8\ \text{N/kg}$ , to a location where the field strength is $1.6\ \text{N/kg}$ . Find the toolbox's mass and weight in both places, and find how much force a person must exert to hold it still in each place. Take up as the positive direction.

Set the convention first: positive $y$ is up, so the weight is negative and any hold-up force is positive.
Mass on Earth: $6.0\ \text{kg}$ . This is given, and no calculation converts it into anything else.
Weight on Earth: $F_g = mg = (6.0\ \text{kg})(9.8\ \text{N/kg}) = 58.8\ \text{N}$ , directed downward. Note the units multiply cleanly, kilograms times newtons per kilogram leaves newtons.
Mass in the weaker field: still $6.0\ \text{kg}$ . Nothing was done to the toolbox, and EK 2.6.D.1 makes mass a property of the object rather than of its surroundings.
Weight in the weaker field: $F_g = (6.0\ \text{kg})(1.6\ \text{N/kg}) = 9.6\ \text{N}$ , downward. The ratio of the two weights is $58.8 / 9.6 = 6.125$ , which is exactly the ratio of the two field strengths, $9.8 / 1.6 = 6.125$ , because the mass cancels out of the ratio.
Hold it still on Earth. At rest, $a_y = 0$ , so $F_{\text{hold}} - 58.8\ \text{N} = 0$ and $F_{\text{hold}} = 58.8\ \text{N}$ upward. In the weaker field the same reasoning gives $9.6\ \text{N}$ upward.
Now a check that separates the two ideas. To hold the toolbox up you need 6.125 times as much force on Earth. To shove it sideways from rest to $2.0\ \text{m/s}$ in $1.0\ \text{s}$ you need $F = ma = (6.0)(2.0) = 12\ \text{N}$ in both places, because that calculation uses the mass and never touches $g$ .

Mass 6.0 kg in both places. Weight 58.8 N on Earth and 9.6 N in the weaker field, both downward, and the holding force matches the weight in each case at $58.8\ \text{N}$ and $9.6\ \text{N}$ upward. The horizontal shove needs $12\ \text{N}$ in either location. Lifting is a weight problem and depends on where you are; accelerating sideways is a mass problem and does not.

Weight on a planet you have never visited

A planet has mass $6.4 \times 10^{23}\ \text{kg}$ and radius $3.4 \times 10^{6}\ \text{m}$ . Using $G = 6.67 \times 10^{-11}\ \text{N} \cdot \text{m}^2/\text{kg}^2$ , find the gravitational field strength at its surface, then find the weight there of the same 6.0 kg toolbox and compare it with the toolbox's weight on Earth.

Start from the printed law $\lvert \vec{F}_g \rvert = \frac{G m_1 m_2}{r^2}$ and divide by the small object's mass to get the field strength, which is the derived relation under EK 2.6.A.2.i: $\lvert \vec{g} \rvert = \frac{GM}{r^2}$ .
Numerator: $GM = (6.67 \times 10^{-11})(6.4 \times 10^{23}) = 4.2688 \times 10^{13}$ .
Denominator: $r^2 = (3.4 \times 10^{6})^2 = 1.156 \times 10^{13}$ .
Divide: $g = \frac{4.2688 \times 10^{13}}{1.156 \times 10^{13}} = 3.693\ \text{N/kg}$ , which rounds to $3.7\ \text{N/kg}$ at two significant figures, matching the two significant figures in the given mass and radius.
Weight of the toolbox there: $F_g = mg = (6.0\ \text{kg})(3.693\ \text{N/kg}) = 22.2\ \text{N}$ , so $22\ \text{N}$ to two significant figures.
Compare with Earth: $\frac{3.693}{9.8} = 0.377$ , so the toolbox weighs about 38 percent as much there. Its mass is $6.0\ \text{kg}$ in both places, and no step of this calculation could have changed that, because the mass was an input on both sides.
Sanity check on the algebra. Notice that the toolbox's own mass cancelled out of the field-strength calculation entirely: $g$ depends on the planet's mass and radius only. That is why one field strength serves every object at that surface, and why weight is proportional to mass at any one location.

Field strength $g = 3.7\ \text{N/kg}$ at that surface. The 6.0 kg toolbox weighs 22 N there against 58.8 N on Earth, about 38 percent as much, while its mass is 6.0 kg in both places. The planet's own mass and radius set $g$ ; the toolbox's mass then sets its weight.

A beam balance and a spring scale disagree

A 2.0 kg bag of flour is weighed two ways on Earth, where the field strength is $9.8\ \text{N/kg}$ : on a two-pan beam balance against standard masses, and on a spring scale whose dial is printed in kilograms. Both instruments are then carried to a location where the field strength is $1.6\ \text{N/kg}$ and used again. Find all four readings and say which instrument measures mass.

On Earth, the spring scale responds to the gravitational force on the bag: $F_g = (2.0\ \text{kg})(9.8\ \text{N/kg}) = 19.6\ \text{N}$ . Its dial was printed by dividing the force it feels by $9.8\ \text{N/kg}$ , so it displays $19.6 / 9.8 = 2.0\ \text{kg}$ . Correct, and correct by coincidence of calibration.
On Earth, the beam balance puts standard masses on the other pan until the beam levels. It levels when the two gravitational forces match: $m_{\text{flour}}\, g = m_{\text{standards}}\, g$ . The $g$ appears on both sides and cancels, so the balance levels at $m_{\text{standards}} = 2.0\ \text{kg}$ .
In the weaker field, the spring scale now feels $F_g = (2.0\ \text{kg})(1.6\ \text{N/kg}) = 3.2\ \text{N}$ . Its dial still divides by the 9.8 that was printed into it, so it displays $3.2 / 9.8 = 0.33\ \text{kg}$ .
In the weaker field, the beam balance still levels at $2.0\ \text{kg}$ . The cancellation in step two did not depend on the value of $g$ , only on it being the same on both pans.
Collect the four readings. Spring scale: $2.0\ \text{kg}$ then $0.33\ \text{kg}$ , wrong by a factor of $9.8 / 1.6 = 6.125$ in the second location. Beam balance: $2.0\ \text{kg}$ then $2.0\ \text{kg}$ .
Name what each device did. The beam balance compared two gravitational forces and let $g$ divide out, so it measured mass. The spring scale measured a single force, in newtons, and then applied a conversion that is only valid in one field. A spring scale marked in kilograms is a force meter with a misleading dial.

Beam balance: 2.0 kg in both locations. Spring scale: 2.0 kg on Earth and 0.33 kg in the weaker field, an error of a factor of 6.125 caused entirely by the 9.8 baked into its calibration. The balance measures mass because $g$ cancels between the two pans; the spring scale measures a force and only reports a correct mass where the field strength matches its calibration.

Frequently asked questions

What is the difference between mass and weight?

Mass is a property of an object itself: how much it resists being accelerated, and how strongly it takes part in gravitational attraction. It is a scalar measured in kilograms and it is the same number everywhere. Weight is the gravitational force that an astronomical body exerts on the object, which the AP Physics 1 CED defines at essential knowledge 2.6.A.3. It is a vector measured in newtons, it points toward the center of the attracting body, and its size is the mass times the local gravitational field strength. Move an object to a place where the field is weaker and its mass is unchanged while its weight is smaller.

Does your mass change on the Moon?

No. Your mass is the same on the Moon, on Earth, in orbit and in deep space, because mass is a property of the matter you are made of rather than of your surroundings. What changes is your weight, because weight is mass times the local gravitational field strength and the Moon's surface field is weaker than Earth's. A beam balance would give the same reading in both places, since it compares two gravitational forces and the field strength cancels between the pans. A spring scale marked in kilograms would give a smaller reading, because it is really measuring a force and dividing by the value of g it was calibrated with on Earth.

Is weight measured in kilograms or newtons?

Newtons. Weight is a force, and the SI unit of force is the newton. Kilograms measure mass. The confusion comes from bathroom scales, which respond to a force but print the result in kilograms after silently dividing by 9.8 newtons per kilogram, a conversion that is only valid where the gravitational field strength is that value and nothing is accelerating. On the AP Physics 1 equation sheet the unit-symbols table lists kilogram and newton as separate entries, and the symbol key uses the letter W for work rather than for weight. Writing a weight in kilograms on an exam answers a different question from the one asked.

Why do heavy and light objects fall at the same rate if the heavy one weighs more?

Because the same mass appears on both sides of the calculation and cancels. The gravitational force on an object is proportional to its mass, and the acceleration that force produces is inversely proportional to its mass, so the two effects exactly offset and every object at the same location falls with the same acceleration. That cancellation is not a trick of algebra: it works only because an object's gravitational mass and its inertial mass are the same number, which essential knowledge 2.6.D.3 records as an experimentally verified equivalence. Essential knowledge 2.6.A.2.ii states the consequence directly, that when gravity is the only force acting, the acceleration in metres per second squared is numerically equal to the field strength in newtons per kilogram.

Are astronauts in orbit weightless?

They appear weightless, which is not the same thing. In low orbit the gravitational field is only a little weaker than at Earth's surface, so gravity is still pulling on an astronaut with most of the force it would exert on the ground. What has gone to zero is the supporting force, because the astronaut and the spacecraft are falling together and neither can push on the other. The AP Physics 1 CED words this carefully at essential knowledge 2.6.C.3: a system appears weightless when there are no forces exerted on the system or when the force of gravity is the only force exerted on the system. Essential knowledge 2.6.C.1 identifies the reading a scale would give with the magnitude of the normal force, which is zero here.

What is the difference between inertial mass and gravitational mass?

Inertial mass is how strongly an object resists a change in its velocity, and it is the mass in Newton's second law. Gravitational mass is how strongly an object takes part in gravitational attraction, and it is the mass in the gravitational force law. The AP Physics 1 CED defines them separately, at essential knowledge 2.6.D.1 and 2.6.D.2, because they answer different questions and nothing in the theory requires them to match. Experiments say they do match, which essential knowledge 2.6.D.3 states as an equivalence, and that is what makes free-fall acceleration independent of the falling object. In every calculation you will do, one number serves both roles.

Is weight a vector or a scalar?

Weight is a vector, because it is a force. It has a magnitude, mass times the gravitational field strength, and a direction, toward the center of mass of the attracting body, which near Earth's surface means straight down. The AP Physics 1 equation sheet writes the gravitational force with an arrow and magnitude bars for exactly this reason. Mass is a scalar: it has a size and a unit and no direction at all, and a negative mass is not a mass pointing the other way. This is one more reason the two cannot be the same quantity in different units.