I need help understanding the concept of true weight vs apparent weight. I understand this much: if someone is standing in an elevator on a scale, the further up they go the less the reading on the scale becomes. But why is this? Is it that distance affects the force of gravity? The further away the object goes [from the Earth's surface] the less the attractive force? Also, if on some other planet with radius $r$ an object is some distance $d$ away from the surface and is 1% less than its true weight on surface, what is the ratio $d/r$?

  • $\begingroup$ Where did you get your definitions? Strictly speaking, "weight" is just mass times force. You could define an "apparent weight" as $mass * \frac{localforce}{earthgravityforce}$ I suppose. $\endgroup$ Sep 24, 2014 at 14:03
  • $\begingroup$ This is a concept sometimes taught at the US freshman physics level. For example, see University Physics, Young & Freeman: books.google.com/… . $\endgroup$ Sep 25, 2014 at 11:46
  • $\begingroup$ Carl Witthoft - where did you get your definition of weight? Mass times force makes no sense. If anything, weight is a special application of F=ma, or in this case, W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. $\endgroup$
    – user16622
    May 31, 2016 at 9:28

8 Answers 8


Okay. Firstly, I would like to point out that you are mixing two very different concepts here:
(1) Variation in the value of gravity $g$ as the distance from the surface of the earth changes.
(2) True and apparent weight

(1) Variation in the value of gravity
Alright. Variation in gravity. Firstly, lets get clear on the value of $g$. What exactly is $g$? It's like this: Suppose you are somewhere. Maybe sitting somewhere having pizza or flying in the sky. The earth applies a force on you. Let's call this force $F$. Then the value of $g$ is simply defined as $F/m$. That's it.
Now suppose the radius of earth is $R$ and you are at distance $d$ from the surface. (Note, from surface of the earth, not the center.) The force applied on you by the earth is $$F = \cfrac{GM_em}{(R+d)^2}$$ So, now, $$g = F/m = \cfrac{GM_e}{(R+d)^2}$$ Have a look at it. The value of $g$ indeed depends on $d$, your distance from the surface of the earth. But, near the surface of the earth, $d<<R$, so we can approximate the above expression to $$g = F/m = \cfrac{GM_e}{R^2}$$ which is independent of $d$. But note that it is valid only for small values of $d$.

(2) True and apparent weight
Okay. Answer to the next part of the question. True and apparent weight. True weight is simply weight. What is your true weight? It's simply $mg$. Mass multiplied by gravity. End of story.
Now, Apparent weight. I'll denote it by $W_A$. It's defined as $$W_A = N$$ where $N$ is the normal force in the direction opposite to the direction of gravity. That is away from the center of the earth. You may be standing and someone may be trying to push you horizontally. That normal reaction force doesn't count. Only the vertical Normal Force counts.
So suppose you jump from the top of the building because your dog died. You are falling. Your '(True) Weight' is simply $mg$. Your Apparent weight is $0$. Because there is no normal force applied on you currently. (Offcourse the ground will apply one hell of a normal force when you finally reach it.)
Now suppose you are standing in an elevator at rest. True weight, offcourse is $mg$. But Apparent weight is also $mg$. Because you are at rest, $N = mg$.
Elevator moving with constant speed: $N = mg$
Suppose the magnitude of elevator's acceleration is $|a|$.
Elevator moving upwards, and slowing down: $N = mg - m|a|$
Elevator moving upwards, and increasing speed: $N = mg + m|a|$
Elevator moving downwards, and slowing down: $N = mg + m|a|$
Elevator moving downwards, and increasing speed: $N = mg - m|a|$
So, why did they introduce the concept of Apparent Weight. Apparent weight is the weight you 'feel'. Think about it! When you are falling, you feel weightlessness. Hence Apparent Weight is $0$. When in an elevator with moving upwards with increasing speed, you feel heavier. Hence more is the Apparent Weight!

(3) $d/R$ ratio
The ratio $d/R$ where weight would be 1% lesser:
$$\cfrac{GM_e}{(R+d_1)^2} = 0.99 \cfrac{GM_e}{R^2}$$ Solve it for $d_1$. That's your answer!

  • $\begingroup$ You are ignoring that $mg$ is NOT true weight. It's apparent weight! The value of $g$, 9.80665 m/s$^2$, is the average apparent acceleration at sea level, at about 45 degrees latitude. That value includes centrifugal acceleration term (the centrifugal acceleration at 45 degrees latitude). $\endgroup$ Sep 24, 2014 at 17:00
  • 3
    $\begingroup$ @DavidHammen THAT value of $g$ is for VERY specific conditions. In my entire answer, by $g$, I mean, 'the acceleration due to gravity at that place'. I don't mean that standard value of $g$ define by some commitee. And you are wrong. $mg$ IS the true weight. $\endgroup$
    – John Red
    Sep 24, 2014 at 17:06

The astronauts on cosmonauts on the International Space Station exhibit a marked difference in their "true" and "apparent" weights. Their true weight, tautologically mass times gravitational acceleration, is about 10% less than what it is on the surface of the Earth. Their apparent weight is essentially zero.

Except at the poles, there's a slight difference between true and apparent weight of an object sitting still on the surface of the Earth. Consider an object sitting on a scale the equator. The forces on the object are the upward normal force exerted on the object by the scale and the downward force of gravity exerted on the object by the Earth as a whole. The object is rotating with the Earth, so it is undergoing uniform circular motion, one revolution per sidereal day (about 7.292116×10-5 s-1), at a distance of 6378.137 km from the center of the Earth. This means the net force on the object cannot be not zero. (It's about 2.5 newtons for a 74 kg object.) Since the net force is not zero, the true and apparent weights cannot be the same.

The difference between true and apparent weight from a Newtonian perspective is that true weight is the magnitude of the force due to gravity, $W_{\text{true}} = \frac {GMm}r^2$ for a small test mass of mass $m$ attracted gravitationally to an object with mass $M$ and a spherical mass distribution. Apparent weight is the magnitude of the sum of all real net forces except for gravity: $W_{\text{apparent}} = \left| \sum \mathbf F \right|$, where the sum is taken over all real, non-gravitational forces acting on the body in question.

From a general relativistic perspective, the concept of "true" weight has little meaning. The concept of "apparent weight" does. It's the magnitude of the net real force acting on an object. Gravitation isn't a real force in general relativity, so there's no reason for that "real, non-gravitational". All that's needed is "real".

Most of the answers are even more confused than the person who asked the question. The following table depicts apparent and true weight of a person massing 75 kg.

$$ \begin{matrix} \text{Location} & \text{Apparent weight}\,(\text{N}) & \text{True weight}\,(\text{N}) \\ \text{North pole} & 737.53 & 737.53 \\ \text{Equator, sea level} & 733.52 & 730.98 \\ \text{Nevado Huascarán, peak} & 732.29 & 729.78 \\ \text{Shock Wave roller coaster, 5.9 g} & 4340 & 735.50 \\ \text{Vomit Comet, top of arc} & 0 & 735.50 \\ \text{Space Station, 330 km altitude} & 0 & 664.37 \end{matrix} $$

These situations depict a number of the ways in which true weight and apparent weight differ, and how the differences between the two can be quite marked.

Ignoring buoyancy and tidal forces, the apparent and true weight of a person at the North Pole are one and the same. At the equator, the person's true weight is reduced from the polar value because the person is about 21 km further from the center of the Earth. The apparent weight is reduced even further because the person is rotating with the Earth. The peak of Nevado Huascarán is where the surface value of apparent gravitational acceleration reaches its minimum. This is partly because it's close to the Equator, but also partly because mountains are huge masses of less dense material floating on the more dense lithosphere.

The last three cases represent what happens with moving objects. That roller coaster ride exerts very strong g forces on the riders, up to about 5.9 g. The Vomit Comet was an airplane NASA used to accommodate astronauts to a zero g environment. NASA now contracts this work out; individuals can now buy tickets and feel what weightlessness feels like. The true weights on the roller coaster and Vomit Comet are more or less the same. The apparent weights differ markedly. Finally, the International Space Station orbits at about 330 km above the Earth's surface. This reduces true weight to about 90% of the surface value. The apparent weight? It's zero, the same as the value at the top of the arc of a Vomit Comet ride.


Let us first understand this concept with a more fundamental situation.

Suppose you are falling from a 50 storey building. If you care to, you'll feel that you're weightless during the fall. You know that gravity is acting on you, but why do you 'feel' that no force is acting on you ? The reason is that although gravity does act on you, there is no upward (normal) force on your feet to oppose the force of gravity. Therefore, you don't feel 'compressed' due to the upward force on your feet and the downward force of gravity as there is no upward force. Also, since your height is negligible compared to the dimensions of the earth, the acceleration of every part of your body is same as every other part; meaning you don't feel 'stretched' either. Remember that all the forces I'm describing are with respect to you.

Your 'apparent weight' in this case is zero. That is, the weight you 'feel' is zero. While standing, the normal force on you is exactly equal to the force due to gravity (you're fully 'compressed'); therefore, you feel your 'whole' weight. That is, the apparent weight is equal to your actual weight.

Now, in the elevator accelerating downwards (say), the very reason you are accelerating downwards is because the normal force is lesser than it is when you're standing and therefore, the constant gravitational force can overcome it. Thus, now your 'apparent' weight is lesser than your actual weight. It is now equal to m(g-a), where 'a' is your downward acceleration. Hope it helped!

P.S: The answer I've given here is just a intuitive explanation of the apparent weight concept. I think you can grasp the quantitative or the mathematical part from any of the comments here or from any good physics textbook.


Let's try to make the answer as simple as possible.

Static weight is written as $w=mg$.(Notice Newtons 2nd law looks similar $F=ma$.) This is $$\mathrm{mass} \times \mathrm{one\ unit\ of\ gravity}$$. On earth, $g=9.8\frac{m}{s^2}$. On a smaller planet, $g$ is less. On a larger planet, $g$ is more. It is important to understand that despite acceleration in the equation, this formula for weight applied to motionless and/or constant velocity. If there is no change in velocity, there is no net force.

Apparent weight can refer to different circumstances:

  1. If a mass is submerged in a fluid, i.e. swimming pool your apparent weight is less.
  2. If you are a great distance from the center of earth, i.e. near edge of atmosphere.
  3. This is the most common reference. It involves the change in weight (downward force) when there is is a change in velocity in the vertical direction. A common example is the elevator example. If you stand on a scale in an elevator, your weight will appear, hence term apparent weight. An example: Get on an elevator and stand on a scale. Your weight will be same as in your bathroom. When the elevator accelerates upward, your weight will increase. When it reaches its max velocity your weight will return to same as static weight, since there is no change in velocity. As it approaches your upper floor it will obviously have to slow down, and your weight will appear less than static weight. This is an exact science, if you weight 200 lbs, and your apparent weight is 240, you can be certain that the acceleration of the elevator is 20% of $9.8\frac{m}{s^2}$. The inverse is also true. If you enter the elevator on the 100th floor, and press ground floor, your weight will appear less until the elevator reaches its max velocity. I.e. if same 200lb man is on scale and weight appears to be 120 lbs(60% of 200), the elevator is accelerating at 40% of g. If the elevator was to accelerate at $9.8\frac{m}{s^2}$, which it wouldn't, your apparent weight will be 0. An object in freefall can do no work because it is not exerting any downward Force.

The only way a mass can exert an amplified Force, acting by gravity alone, is if it decelerates.

For a real world meaningful application of this principle we can look at the tragic collapse of the WTC of 9/11. A frame by frame analysis of the top section of the building shows that the top 12 floor section accelerates directly through what should be its collision with the stronger, undamaged, progressively stronger 95 floors at a rate of $6.41\frac{m}{s^2}$. So it is exerting only about 1/3 the Force if it were simply sitting motionless. Since the building was engineered to handle 3-5 time the load, we know the top section is not crushing the lower section. The lower section is being destroyed by some force allowing top section to fall through the lower section.


So, I assume by "apparent weight" you mean the reading on the scales, compared with a "true weight" which is the reading on the scales when the lift is stationary on the ground floor?

Right, so first let's nail a misconception. The reading on the scales is not going to become less the higher you go, not unless you are in a very tall building indeed or have a very accurate set of scales. The radius of the Earth is about 6400km; gravitational field strength depends on $1/r^2$, so even increasing your height to $d=829$m (the Burj Khalif in Dubai), decreases gravity by a factor of $r^2/(d+r)^2 = 0.99974$ - measurable, certainly, but possibly not on a set of bathroom scales. And I think this shows you how to answer the last part of your question.

No, the effect you are thinking of occurs when the lift is accelerating or decelerating at the beginning and end of its journey up the building. As the lift accelerates, this acceleration needs to be added to the acceleration due to gravity (about 9.81 m/s$^2$). This total acceleration multiplied by your mass (your mass is your "true weight"/9.81) gives you your (heavier) "apparent weight".

In similar fashion, at the end of the lift journey, the lift decelerates and this deceleration is subtracted from the acceleration due to gravity and the total is multiplied by your mass to give the (lighter) "apparent weight".

This effect is readily apparent - you can feel it happening in most lifts. A typical acceleration/deceleration is of order 1 m/s$^2$.

Here is a nice demonstration, if you can get it to run.

See also this related question Accelerating an elevator

There are also plenty of other resources all over the internet - it's a fairly standard question.


Unlike the other answers I'm going to assume that you have been given a hyperbolic example of a building that reaches up into space. What you would call the 'true weight', I suppose, is the reading of a set of scales on the surface of the Earth, if these scales are calibrated for the surface of the Earth. These actually measure the gravitational force acting on you: $$ F = G \frac{Mm}{r^2} $$ Here $M$ is the mass of the Earth, $r$ is the distance from the centre of the Earth, $G$ is the gravitational constant, and $m$ is your mass. Let's call the radius of the Earth $R$, so that the weight on Earth's surface is $F_{s} = GMm/R^2$.

If you move up high to some new distance $d$ above the surface, the force acting on you will be $$ F_d = G \frac{Mm}{(R+d)^2} $$

To get a measure of the apparent weight, you just need to check the ratio $A$ of these weights, which will be a function of the distance you are from the surface:

$$ A(d) = \frac{F_d}{F_s} = \frac{R^2}{(R+d)^2} $$

You can just as well apply this ratio to the measured mass to find the true mass if you're actually using bathroom scales. So if your weight is $F_s$, your apparent weight at height $d$ will be

$$ F_d = A(d)\ F_s$$

or, in terms of 'apparent mass', i.e. the reading on a scale that's calibrated for the surface of the Earth,

$$m_{\mathrm{apparent}} = A(d)\ m $$

Note that if $d$ is much smaller than the radius of the Earth, then $A$ will be very close to 1, which is why we don't notice this effect in our daily lives. For example, at a height of 100km, assuming the Earth's radius to be 6400km, we have $A(100\mathrm{km}) = 0.969...$, so even at that height your apparent weight will still be almost 97% of your regular weight.

I'll leave it up to you to invert the relation and find $d$ in terms of $A$.

  • $\begingroup$ This answer is wrong. Bathroom scales measure apparent weight rather than true weight. Balance scales measure mass rather than weight. There is no local experiment that can measure true weight. $\endgroup$ Sep 24, 2014 at 16:09
  • $\begingroup$ @DavidHammen That's why I insisted that they be calibrated to the surface of the Earth. Though I admit I didn't mention that until later in the comment, I'll move it higher up. $\endgroup$
    – johnpaton
    Sep 24, 2014 at 16:42
  • $\begingroup$ This answer remains confused. Variations in gravitational force with altitude have nothing to do with apparent versus true weight. Your true weight at the top of a very tall mountain is less than your true weight at sea level at the same latitude as the mountain. $\endgroup$ Sep 24, 2014 at 18:26
  • $\begingroup$ @DavidHammen I think this is more an issue of interpreting the question. I assumed he was essentially just interested in the inverse-square law, whereas you are taking accelerations into account. Without clarification from OP it's impossible to know what (s)he's looking for. $\endgroup$
    – johnpaton
    Sep 25, 2014 at 11:26
  • $\begingroup$ This is a matter of commonly used definitions. For example, see books.google.com/… . $\endgroup$ Sep 25, 2014 at 11:41

While it is true that the gravitational force dissipates with respect to distance squared, that is not the reason a scale would output a "different weight". A scale does not actually measure weight, only it's response to it. That is, the scale reads the normal force. If the elevator was motionless, the normal force would be equivalent in magnitude to your weight. Therefore, the scale reading would also happen to be your weight.

Now consider an accelerating elevator. We can easily analyze what's going on mathematically. (Normal Force) - (Your Weight) = (Your mass) * (acceleration).

 Normal Force = (Your mass * acceleration) + (Your weight). 

That is the normal force is the sum of your weight and the relative force associated with the accelerating elevator.

Physically, you can think about the electrons (which account for the normal force) in the scale. When the elevator is motionless (or constant speed), and you are standing on the scale, those electrons will push back with equal force. However, when the elevator is accelerating upward, those electrons are forced to be closer to your feet. The response then is simply a larger normal force.


True weight actually the product of mass and gravitational acceleration which is equal to mg where the apparent weight is the sum of net forces ( when you standing in elevator and elevator is moving either upwards or downwards, either high speed or low speed then you feel your weight heavier or lighter this is the apparent weight that u feels which is equal to sum of net forces).On the other hand when you jump from a certain height you feel weightless in that time no normal force present then net force 0 so that time apparent weight is zero. In short apparent weight is the weight that you feel . Mathematically you can find the apparent and tue weight of an object which is F(d)=F(s) A(d) Here F(s) is your real weight and F(d) is apparent weight and A(d) is the ratio of force of gravity on earth surface and force of gravity at height [R^2/(R+d)^2]


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