Gravitational Field vs Force: The Difference

A gravitational force is an interaction between two objects with mass, measured in newtons. A gravitational field is a property of a point in space, measured in newtons per kilogram, and it is there whether or not an object is. Divide the force by the test object's mass to get the field.

AP Physics: Unit 2 (topics 2.6 Gravitational Force, 10.3 Electric Fields). Both quantities are defined in Topic 2.6, Gravitational Force, inside AP Physics 1 Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. EK 2.6.A.1 gives the force law as a relevant equation, with 2.6.A.1.i stating that the gravitational force is attractive, 2.6.A.1.ii that it is exerted along the line connecting the centers of mass of the two interacting systems, and 2.6.A.1.iii that it can be considered to be exerted on the system's center of mass. EK 2.6.A.2 introduces the field as a model of the effects of a noncontact force exerted on an object at various positions in space, and EK 2.6.A.2.i defines its magnitude as the ratio of the gravitational force on a test object to that object's mass, with the derived equation g = F_g/m = GM/r squared. EK 2.6.A.2.ii states that when the gravitational force is the only force exerted on an object, the acceleration in metres per second squared is numerically equal to the field strength in newtons per kilogram, which is why the constants box of all four booklets prints g = 9.8 m/s squared and g = 9.8 N/kg on separate lines. EK 2.6.A.3 defines weight as the gravitational force exerted by an astronomical body on a relatively small nearby object. EK 2.6.B.2 gives g approximately 10 N/kg near Earth's surface, and EK 2.6.C.1 through 2.6.C.4 cover apparent weight and the equivalence principle. Suggested skills for Topic 2.6 are 1.A, 2.A, 2.D and 3.C in AP Physics 1 and 1.C, 2.A, 2.D and 3.B in AP Physics C: Mechanics, which adds LO 2.6.E on uniform spherical distributions of mass under a boundary statement that students are not expected to mathematically prove or derive Newton's shell theorem. The electric analogue is EK 10.3.A.2 in AP Physics 2 and Topic 8.3, Electric Fields, in AP Physics C: Electricity and Magnetism.

One needs two objects, the other needs one

A gravitational force is something two objects do to each other. A gravitational field is something one object does to space. That is the distinction, and every other difference on this page follows from it.

The AP Physics 1 CED introduces them in that order in Topic 2.6, Gravitational Force.

EK 2.6.A.1 gives the force: Newton's law of universal gravitation describes the gravitational force between two objects or systems as directly proportional to each of their masses and inversely proportional to the square of the distance between the systems' centers of mass, with the relevant equation $\lvert \vec{F}_g \rvert = \dfrac{G m_1 m_2}{r^2}$ . Two masses appear in that equation. Delete either and there is no force.

EK 2.6.A.2 then introduces the field, and the sentence is doing real work: a field models the effects of a noncontact force exerted on an object at various positions in space. Not at a position; at various positions. A field is a map, and a map does not need a traveller.

EK 2.6.A.2.i makes the definition operational: the magnitude of the gravitational field created by a system of mass $M$ at a point in space is equal to the ratio of the gravitational force exerted by the system on a test object of mass $m$ to the mass of the test object. Its derived equation is the whole comparison in one line:

\lvert \vec{g} \rvert = \frac{\lvert \vec{F}_g \rvert}{m} = G\frac{M}{r^2}

Read the two halves. The middle expression says how you measure a field: put something there, divide out its mass. The right-hand expression says the answer does not contain $m$ , so the thing you measured belongs to $M$ and to the point, not to whatever you used to probe it. The test object is scaffolding. The field is what is left after you take the scaffolding away.

This is the parent of the electric case. EK 10.3.A.2 in AP Physics 2 defines the electric field at a given point as the ratio of the electric force exerted on a test charge at that point to the charge of the test charge, with the equation $\vec{E} = \vec{F}_E / q$ , and EK 10.3.A.2.i defines a test charge as one of small enough magnitude that its presence does not significantly affect the field in its vicinity. Same structure, same word, same division.

Gravitational field vs gravitational force, side by side

Question you are asking	Gravitational force $\vec{F}_g$	Gravitational field $\vec{g}$
CED essential knowledge	2.6.A.1	2.6.A.2 and 2.6.A.2.i
Equation	$\lvert \vec{F}_g \rvert = \dfrac{G m_1 m_2}{r^2}$	$\lvert \vec{g} \rvert = \dfrac{\lvert \vec{F}_g \rvert}{m} = G\dfrac{M}{r^2}$
How many objects it takes	Two	One, the source
Whose property it is	The interaction	The point in space
SI unit	$\text{N}$	$\text{N/kg}$ , numerically also $\text{m/s}^2$
Scalar or vector	Vector	Vector
Depends on the test object's mass	Yes, directly proportional	No
Exists with nothing there to feel it	No	Yes
Direction	Attractive, along the line joining the centers of mass	Toward the source mass
What doubles it	Doubling either mass	Doubling the source mass only
Effect of doubling $r$	Falls to one quarter	Falls to one quarter
Printed in the constants box	Not applicable	$g = 9.8\ \text{N/kg}$ at Earth's surface
Its electric counterpart	Coulomb's law	$\vec{E} = \vec{F}_E / q$

Four rows carry most of the weight.

The unit row is the distinction encoded in symbols. A newton per kilogram is a force with the receiving object's mass already divided out. Any quantity whose unit has been divided by a property of the test object cannot be a property of the test object.

The dependence row is what makes the field useful. Put a $2$ kg object and a $60$ kg object at the same point above Earth and they feel forces that differ by a factor of $30$ . They sit in one field. That is why a field is worth defining: it is the part of the situation that does not change when you change what you are measuring with.

The row about existing with nothing there is the conceptual core, and the reason EK 2.6.A.2 says at various positions in space. The field above a lecture bench is a definite number at every point in the room right now, and it was that number before anyone walked in.

The direction row hides an asymmetry worth noticing. The gravitational force is a mutual pair: EK 2.6.A.1.ii says it is always exerted along the line connecting the centers of mass of the two interacting systems, and by Newton's third law each object pulls the other equally hard. The field is not mutual. It points one way, from the point toward the source, because it was defined by dividing out one of the two objects.

The case that separates them: two test objects at one point

Hold a $2.0$ kg textbook and a $60$ kg person at the same place just above Earth's surface, where the field is $9.8$ N/kg.

Quantity	Textbook	Person
Mass	$2.0$ kg	$60$ kg
Gravitational force on it	$19.6$ N	$589$ N
Gravitational field at its location	$9.8$ N/kg	$9.8$ N/kg
Acceleration if released	$9.8\ \text{m/s}^2$	$9.8\ \text{m/s}^2$

Two different forces and one field. The force column tells you about the objects. The field column tells you about the place. Worked example one runs the numbers from the planet's mass and radius rather than assuming $9.8$ .

Then do the experiment that makes the point unanswerable: take both objects away. The forces become zero, because a force between two objects needs two objects. The field stays at $9.8$ N/kg, because Earth is still there and the point is still there. If the field were only a way of talking about forces, it would have vanished with them.

The superposition case is where the split earns its keep. Suppose a planet of mass $6.0 \times 10^{24}$ kg and a moon of mass $1.5 \times 10^{24}$ kg have centers $4.0 \times 10^8$ m apart. Between them is a point where the net field is exactly zero, and worked example three locates it at two thirds of the way from planet to moon. At that point a spacecraft feels no net gravitational force while being pulled by two forces of about $6.75$ N each. The field vanishes; the forces do not, they cancel. Trying to describe that situation with forces alone means tracking two vectors for every object you put there. The field does the cancelling once, at the point, for everything that will ever pass through it.

One more separation, this one about numbers rather than concepts. Two masses can produce the same force at different places or different forces at the same place, and the field is the bookkeeping that tells you which. Change the test object and the force changes and the field does not. Change the point and both change. Change the source and both change. Only one of the three changes leaves the field alone, and it is the one people most often think should move it.

Why g is printed in two different units, and why both are right

Look at the constants box on any of the four AP Physics equation sheets and you will find $g$ twice, on two lines, with two units:

Magnitude of the acceleration due to gravity at Earth's surface, $g = 9.8\ \text{m/s}^2$
Magnitude of the gravitational field strength at Earth's surface, $g = 9.8\ \text{N/kg}$

Both lines appear in the constants box of the AP Physics 1, AP Physics 2, AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism booklets, checked against the appendix pages of all four Course and Exam Descriptions. That is not redundancy. It is the field and its consequence, printed as two separate facts because they are two separate facts.

The CED states the connection at EK 2.6.A.2.ii: 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. Read the conditional. The equality holds when gravity is the only force. Hold the object still and the field strength is unchanged while the acceleration is zero.

The units are the same because a newton is a $\text{kg}\cdot\text{m}/\text{s}^2$ :

1\ \frac{\text{N}}{\text{kg}} = \frac{1\ \text{kg}\cdot\text{m}/\text{s}^2}{1\ \text{kg}} = 1\ \frac{\text{m}}{\text{s}^2}

So nothing is being converted. What differs is which question the number answers.

N/kg answers a question about the field: how much force per kilogram is available here. It is defined by EK 2.6.A.2.i as a ratio of a force to a mass, and it is meaningful with no object present.
$\text{m/s}^2$ answers a question about an object: how fast is this thing's velocity changing. It requires an object, and it requires that nothing else is pushing on it.

Writing N/kg is the honest unit when you are describing a location, and $\text{m/s}^2$ is the honest unit when you are describing a fall. A book on a table sits in a field of $9.8$ N/kg and has an acceleration of zero, and the two statements are not in conflict.

A rounding note, because it trips people who read the CED body and the printed table on the same day. EK 2.6.B.2 says that near the surface of Earth the strength of the gravitational field is $g \approx 10$ N/kg, while the printed Table of Information gives $9.8$ . Both are in the same document. Is g 9.8 or 10 works through which number to use where.

Weight, apparent weight, and the reading a scale actually gives

The field is what makes weight definable, and the place people trip is that a bathroom scale does not measure it.

EK 2.6.A.3: the gravitational force exerted by an astronomical body on a relatively small nearby object is called weight. The derived equation is $\text{Weight} = F_g = mg$ . So weight is a force, in newtons, and it is the field at your location multiplied by your mass. It is not a property of you alone, and it changes when you change planets while your mass does not.

What a scale reads is a different quantity. EK 2.6.C.1: the magnitude of the apparent weight of a system is the magnitude of the normal force exerted on the system. And EK 2.6.C.2: if the system is accelerating, the apparent weight of the system is not equal to the magnitude of the gravitational force exerted on the system. A scale in a lift measures the normal force, and that is what changes as the lift accelerates.

This produces a situation that is routinely misdescribed. EK 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. An astronaut in orbit is in the second case, and worked example two puts numbers on it: at an orbital altitude of $400$ km the field is about $8.69$ N/kg, some $89$ percent of its surface value, and the gravitational force on a $60$ kg astronaut there is about $521$ N. The field is nearly as strong as it is on the ground. The apparent weight is zero because nothing is pushing back.

So there are three distinct quantities and only two of them are the field's business.

Quantity	Symbol	What it is	Zero in orbit
Gravitational field	$\vec{g}$	Force per unit mass at the point	No
Weight, that is the gravitational force	$\vec{F}_g$	The field times the object's mass	No
Apparent weight	$\vec{F}_N$	The normal force on the object	Yes

EK 2.6.C.4 closes the loop with 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. That is the CED saying that the two right-hand columns can be confused by an observer for good physical reasons, which is a stronger statement than a warning about sloppy vocabulary.

Where the confusion costs a mark

Each of these is a specific scoring error.

Quoting the field in newtons. $\vec{g}$ is a force per unit mass. A field of $9.8$ N is not a field.
Quoting the force in N/kg. The other half. $\lvert \vec{F}_g \rvert = G m_1 m_2 / r^2$ has both masses in it and comes out in newtons.
Writing $g = 9.8$ into a problem on another planet, or far from Earth. The printed value is labelled at Earth's surface. Off the surface you need $\lvert \vec{g} \rvert = GM/r^2$ , and worked example one shows the field at $400$ km is already down to $8.69$ N/kg.
Measuring $r$ from the surface. In both $GM/r^2$ and $Gm_1m_2/r^2$ , $r$ runs between centers of mass, which EK 2.6.A.1.ii states for the force. An altitude is not an $r$ ; add the planet's radius.
Saying the field is zero in orbit. It is not, and saying so contradicts the very orbit being described, since the field is what curves the path. What is zero is the apparent weight, by EK 2.6.C.3.
Saying an astronaut's weight is zero in orbit. By EK 2.6.A.3 weight is the gravitational force, which is about $521$ N for a $60$ kg astronaut at $400$ km. Apparent weight is the zero one.
Making the field depend on the object placed in it. The whole point of EK 2.6.A.2.i is the division. Two objects of different mass at one point share one field.
Treating the field as mutual. The force is a Newton's third law pair, equal in magnitude on both objects. The field is not: the field of Earth at your location and your own field at Earth's location are wildly different numbers.
Adding fields as scalars. Fields superpose as vectors. Worked example three has a point where two fields of equal magnitude sum to zero, and adding the magnitudes there would give twice the larger one.
Using the field of one body inside another body's interior without care. AP Physics C: Mechanics adds LO 2.6.E for this: EK 2.6.E.2.i says the net gravitational force exerted on an object inside a thin spherical shell is zero, and EK 2.6.E.2.iii says an object inside a sphere of uniform density experiences a net gravitational force from only a partial mass of the sphere. AP Physics 1 does not require this, and its own boundary note is that C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem.

What the CED requires, and where the field idea goes next

Both quantities are defined in one topic, and it is a Unit 2 topic rather than a gravitation unit.

AP Physics 1 Topic 2.6, Gravitational Force, sits in Unit 2, Force and Translational Dynamics, weighted at 18 to 23 percent of the multiple-choice section over about 22 to 27 class periods. The topic carries four learning objectives:

2.6.A, describe the gravitational interaction between two objects or systems with mass, carrying EK 2.6.A.1 with its three sub-statements, EK 2.6.A.2 with its two, and EK 2.6.A.3.
2.6.B, describe situations in which the gravitational force can be considered constant, carrying EK 2.6.B.1 and EK 2.6.B.2.
2.6.C, describe the conditions under which the magnitude of a system's apparent weight is different from the magnitude of the gravitational force exerted on that system, carrying EK 2.6.C.1 through 2.6.C.4.
2.6.D, describe inertial and gravitational mass, carrying EK 2.6.D.1 through 2.6.D.3.

Suggested skills for Topic 2.6 in AP Physics 1 are 1.A, 2.A, 2.D and 3.C.

AP Physics C: Mechanics Topic 2.6 carries all of the above with the same numbering, with suggested skills 1.C, 2.A, 2.D and 3.B, and adds a fifth objective, 2.6.E, describe the gravitational force exerted on an object by a uniform spherical distribution of mass. That objective carries Newton's shell theorem at EK 2.6.E.2 and the interior results at EK 2.6.E.2.i through 2.6.E.2.iv and EK 2.6.E.3, under a boundary statement that AP Physics C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem.

On the sheets, $\lvert \vec{F}_g \rvert = G m_1 m_2 / r^2$ is printed in the mechanics table of all four booklets, and $G = 6.67 \times 10^{-11}$ is in every constants box. The field equation $\lvert \vec{g} \rvert = \lvert \vec{F}_g \rvert / m = GM/r^2$ is labelled a derived equation in the CED rather than a relevant one, and it is not printed on any of the four sheets. That labelling is the CED telling you it expects you to produce it, which for this equation means doing exactly the division EK 2.6.A.2.i describes.

Where the idea goes next is the whole reason it is introduced in Unit 2 rather than alongside orbits. The same construction reappears for charge: in AP Physics 2 at EK 10.3.A.2 with $\vec{E} = \vec{F}_E/q$ , printed on the Physics 2 and C: Electricity and Magnetism sheets, and in AP Physics C: Electricity and Magnetism at Topic 8.3, Electric Fields. EK 10.3.A.3 adds that the electric field is a vector quantity and can be represented in space using vector field maps, which is the picture EK 2.6.A.2's phrase at various positions in space was already describing. Compare the two force laws directly at gravitational vs electric force, and see the CED framing at Topic 2.6 in AP Physics 1, Topic 2.6 in C: Mechanics and Topic 10.3, Electric Fields. The electric field and potential guide carries the electric routines, and the circular motion and gravitation practice set has problems to work.

One field, two test objects, and what altitude does to it

A planet has mass $M = 5.97 \times 10^{24}$ kg and radius $R = 6.37 \times 10^6$ m. Find the gravitational field strength at its surface, and the gravitational force it exerts on a $2.0$ kg textbook and on a $60$ kg person, both at the surface. Then find the field and the force on the same person at an altitude of $400$ km. Use $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ .

Field at the surface, from EK 2.6.A.2.i. $\lvert \vec{g} \rvert = GM/R^2$ . The numerator is $(6.67 \times 10^{-11})(5.97 \times 10^{24}) = 3.982 \times 10^{14}$ . The denominator is $(6.37 \times 10^6)^2 = 4.058 \times 10^{13}\ \text{m}^2$ .
Divide: $\lvert \vec{g} \rvert = (3.982 \times 10^{14})/(4.058 \times 10^{13}) = 9.81\ \text{N/kg}$ , which rounds to the $9.8$ N/kg the constants box prints. Nothing about the test object entered this calculation, because no test object was mentioned.
Force on the textbook. $\lvert \vec{F}_g \rvert = mg = (2.0)(9.81) = 19.6\ \text{N}$ .
Force on the person. $\lvert \vec{F}_g \rvert = (60)(9.81) = 589\ \text{N}$ , that is $5.9 \times 10^2$ N to two figures.
Compare the columns. The two forces differ by a factor of $30$ , exactly the ratio of the masses. The field is one number for both, $9.81$ N/kg. If both objects were removed, both forces would be zero and the field would be unchanged.
Field at $400$ km altitude. The distance in $GM/r^2$ runs from the center, so $r = R + h = 6.37 \times 10^6 + 0.40 \times 10^6 = 6.77 \times 10^6\ \text{m}$ . Then $r^2 = 4.583 \times 10^{13}\ \text{m}^2$ and $\lvert \vec{g} \rvert = (3.982 \times 10^{14})/(4.583 \times 10^{13}) = 8.69\ \text{N/kg}$ .
Force on the person there. $(60)(8.69) = 521\ \text{N}$ .
Read the altitude result. Going up $400$ km, which is about $6$ percent of the planet's radius, cut the field only to $8.69/9.81 = 88.5$ percent of its surface value. An astronaut at that altitude is in a field nearly as strong as on the ground and is pulled with about $521$ N. What is zero up there is the normal force, and therefore the apparent weight, by EK 2.6.C.1 and EK 2.6.C.3.

The surface field is $9.81$ N/kg, and it is the same number for both objects. The forces are $19.6$ N on the $2.0$ kg textbook and $589$ N on the $60$ kg person. At $400$ km altitude the field is $8.69$ N/kg, $88.5$ percent of the surface value, and the force on the $60$ kg person is $521$ N. Neither the field nor the force is zero in orbit; the apparent weight is.

N/kg against m/s squared, on a table and in free fall

A $3.0$ kg block sits at rest on a horizontal table where the gravitational field strength is $9.8$ N/kg. State the field, the gravitational force, the normal force, the apparent weight and the acceleration. Then cut the table away and state all five again. Take down as the positive direction and use $g = 9.8\ \text{m/s}^2$ .

Sign convention, declared before anything else. Down is positive throughout, so the gravitational force is positive and the normal force is negative.
On the table, the field. The field is a property of the location, not of the block or of what is under it: $\lvert \vec{g} \rvert = 9.8\ \text{N/kg}$ . Removing the table will not change it.
On the table, the gravitational force. EK 2.6.A.3 gives $\text{Weight} = F_g = mg = (3.0)(9.8) = +29.4\ \text{N}$ , downward.
On the table, the normal force and apparent weight. The block is in equilibrium, so the normal force balances the gravitational force: $F_N = -29.4\ \text{N}$ , that is $29.4$ N upward. By EK 2.6.C.1 the magnitude of the apparent weight is the magnitude of the normal force, so the apparent weight is $29.4$ N.
On the table, the acceleration. Zero. Note that the field is $9.8$ N/kg and the acceleration is $0\ \text{m/s}^2$ at the same instant, which is only a contradiction if the two units are being treated as the same statement.
Table removed, the field. Still $9.8$ N/kg. Nothing about the source mass or the location changed.
Table removed, the forces. $F_g = +29.4\ \text{N}$ as before. $F_N = 0$ , so the apparent weight is now zero, which is EK 2.6.C.3's condition that the force of gravity is the only force exerted on the system.
Table removed, the acceleration. $a = F_{\text{net}}/m = 29.4/3.0 = +9.8\ \text{m/s}^2$ . Now EK 2.6.A.2.ii applies, because the gravitational force is the only force exerted on the object, and the observed acceleration in $\text{m/s}^2$ is numerically equal to the field strength in N/kg.
The unit identity behind that equality. $1\ \text{N/kg} = 1\ (\text{kg}\cdot\text{m}/\text{s}^2)/\text{kg} = 1\ \text{m/s}^2$ . The numbers were always going to match; what the condition in EK 2.6.A.2.ii decides is whether the acceleration is the one that matches.

On the table: field $9.8$ N/kg, gravitational force $29.4$ N down, normal force $29.4$ N up, apparent weight $29.4$ N, acceleration zero. In free fall: field $9.8$ N/kg, gravitational force $29.4$ N down, normal force zero, apparent weight zero, acceleration $9.8\ \text{m/s}^2$ down. The field and the gravitational force did not change. The apparent weight and the acceleration did.

The point where the field is zero and the forces are not

A planet of mass $6.0 \times 10^{24}$ kg and a moon of mass $1.5 \times 10^{24}$ kg have centers $4.0 \times 10^8$ m apart. Find the point on the line between them where the net gravitational field is zero, and find the two forces exerted on a $1200$ kg spacecraft placed there. Then find the net field at the midpoint between them for comparison. Use $G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2$ .

Set up the condition. Let $x$ be the distance from the planet's center. The two fields point in opposite directions along the line, so the net field is zero where their magnitudes are equal: $\dfrac{GM_1}{x^2} = \dfrac{GM_2}{(d-x)^2}$ . The $G$ cancels, which is the first sign that the answer will not depend on either body's field strength in absolute terms.
Use the mass ratio. $M_1/M_2 = 6.0/1.5 = 4$ , so $\dfrac{4}{x^2} = \dfrac{1}{(d-x)^2}$ , giving $(d-x)^2 = x^2/4$ and therefore $d - x = x/2$ .
Solve. $d = 3x/2$ , so $x = 2d/3 = 2(4.0 \times 10^8)/3 = 2.667 \times 10^8\ \text{m}$ from the planet, leaving $d - x = 1.333 \times 10^8\ \text{m}$ to the moon. The null point is two thirds of the way across, nearer the smaller body, as it must be.
Check by computing both fields. Planet: $(6.67 \times 10^{-11})(6.0 \times 10^{24})/(2.667 \times 10^8)^2 = (4.002 \times 10^{14})/(7.111 \times 10^{16}) = 5.628 \times 10^{-3}\ \text{N/kg}$ . Moon: $(6.67 \times 10^{-11})(1.5 \times 10^{24})/(1.333 \times 10^8)^2 = (1.0005 \times 10^{14})/(1.778 \times 10^{16}) = 5.628 \times 10^{-3}\ \text{N/kg}$ . Equal in magnitude, opposite in direction, so the net field is zero.
Forces on the spacecraft there. Each field acts on the full $1200$ kg: $F = (5.628 \times 10^{-3})(1200) = 6.75\ \text{N}$ toward the planet, and $6.75$ N toward the moon. Neither force is zero, and neither is negligible. The net force is zero because the two cancel, which is a different statement.
Now the midpoint, for contrast. At $x = 2.0 \times 10^8$ m the planet's field is $(4.002 \times 10^{14})/(4.0 \times 10^{16}) = 1.0005 \times 10^{-2}\ \text{N/kg}$ and the moon's is $(1.0005 \times 10^{14})/(4.0 \times 10^{16}) = 2.501 \times 10^{-3}\ \text{N/kg}$ .
Net field at the midpoint. $1.0005 \times 10^{-2} - 2.501 \times 10^{-3} = 7.50 \times 10^{-3}\ \text{N/kg}$ , directed toward the planet. Adding the two magnitudes instead would have given $1.25 \times 10^{-2}$ N/kg, which is the error that treating fields as scalars produces.
Why the field version is the useful one. The null point was found without mentioning any object at all, and the answer applies to every object that will ever pass through it. Doing the same job with forces means recomputing two vectors for each new mass.

The net field is zero at $2.667 \times 10^8$ m from the planet, two thirds of the way to the moon, where each body's field has magnitude $5.628 \times 10^{-3}$ N/kg. A $1200$ kg spacecraft there is pulled by $6.75$ N toward each body, so the forces are far from zero and simply cancel. At the midpoint the net field is $7.50 \times 10^{-3}$ N/kg toward the planet, not the $1.25 \times 10^{-2}$ N/kg that adding magnitudes would give.

Frequently asked questions

What is the difference between gravitational field and gravitational force?

A gravitational force is an interaction between two objects with mass, measured in newtons, and both objects have to exist for it to exist. A gravitational field is a property of a point in space created by one source mass, measured in newtons per kilogram, and it is there whether or not anything is at that point to feel it. The AP Physics 1 CED defines the field at essential knowledge 2.6.A.2.i as the ratio of the gravitational force exerted by the source on a test object to the mass of that test object, which is why the test object's mass divides out and the field belongs to the location rather than to whatever is placed there.

Why is g measured in both m/s squared and N/kg?

Because the two units answer different questions with the same number. A newton per kilogram is a force per unit mass and describes the field at a location, so it is meaningful with nothing there. A metre per second squared is an acceleration and describes an object's motion, so it needs an object and it needs gravity to be the only force acting. The units are numerically identical because a newton is a kilogram metre per second squared, so dividing by a kilogram leaves metres per second squared. All four AP Physics equation sheets print both lines in the constants box: g equals 9.8 metres per second squared for the acceleration due to gravity, and g equals 9.8 newtons per kilogram for the gravitational field strength. Essential knowledge 2.6.A.2.ii states the link, with the condition that the gravitational force is the only force exerted on the object.

Does a gravitational field exist if there is nothing in it?

Yes. That is the point of defining it. Essential knowledge 2.6.A.2 says a field models the effects of a noncontact force exerted on an object at various positions in space, and the phrase at various positions is what makes it a map rather than a single force. The field above your desk has a definite value right now at every point, whether or not anything is there. The test object used to define it at essential knowledge 2.6.A.2.i is scaffolding: divide the force it feels by its own mass, and what is left, G times the source mass over r squared, contains no reference to the test object at all.

Is the gravitational field zero in orbit?

No, and it cannot be, because the field is what curves the orbit. At an altitude of 400 kilometres above a planet of Earth's mass and radius, the field is about 8.69 newtons per kilogram, roughly 89 percent of its surface value, and the gravitational force on a 60 kilogram astronaut there is about 521 newtons. What is zero in orbit is the apparent weight, which essential knowledge 2.6.C.1 defines as the magnitude of the normal force exerted on the system. Essential knowledge 2.6.C.3 states the condition: a system appears weightless when there are no forces exerted on it or when the force of gravity is the only force exerted on it.

How do you calculate gravitational field strength?

Divide the gravitational force on a test object by that object's mass, or equivalently use G times the source mass divided by the square of the distance from the source's center. The AP Physics 1 CED gives both as one derived equation at essential knowledge 2.6.A.2.i. The distance is measured from the center of mass of the source, so an altitude has to have the source's radius added to it before it goes into the formula. Note that this equation is labelled a derived equation rather than a relevant one, and it is not printed on any of the four AP equation sheets, so it has to be produced from the force law and the definition.

Is gravitational field a vector?

Yes. It has a magnitude in newtons per kilogram and a direction, which points from the location toward the source mass, because the gravitational force is attractive by essential knowledge 2.6.A.1.i. Fields from several sources add as vectors, not as numbers, and the difference matters: at a point between a planet and its moon where the two fields are equal in magnitude and opposite in direction, the net field is zero even though each individual field is not. Adding the magnitudes there would give exactly the wrong answer. The electric case is stated explicitly in the AP Physics 2 CED at essential knowledge 10.3.A.3, that the electric field is a vector quantity and can be represented in space using vector field maps.

Is the gravitational field the same as weight?

No. Weight is a force, in newtons, and essential knowledge 2.6.A.3 defines it as the gravitational force exerted by an astronomical body on a relatively small nearby object, with the derived equation weight equals m times g. The field is that force divided by the object's mass, in newtons per kilogram. So weight depends on the object and the field does not: a 2 kilogram book and a 60 kilogram person standing side by side have weights of about 19.6 and 589 newtons in one shared field of 9.8 newtons per kilogram. Apparent weight is a third quantity again, equal to the normal force, and it is the one a bathroom scale reports.