When an object is thrown straight up, its velocity decreases, and at its maximum height the velocity reduces to zero and then it falls back down. My question is, exactly how long does the object stays in the air at zero velocity, and does it (time) varies from place to place due to different gravity?

The object will stay at 0 velocity for an infintensimally small time period (it doesn't last long). To visualize this, let's say for example, the object reaches zero velocity 6 seconds when it is thrown. This means that at 5.9999999999999999 seconds, the object still has some velocity. Only at exactly 6 seconds the ball has 0 m/s. At 6.000000000000000000000000001 seconds, the object has velocity (which is really really really close to zero but not exactly zero). This means that only at that small point of time, exactly at 6 seconds (to infinite precision of digits), it will have exactly 0 m/s. To answer your other question, the time does vary from place to place due to different gravity. For instance, if you would try this same experiment on the Moon, it would take longer for it to fall back since the acceleration due to gravity is slower. (It would also go higher which would mean more distance). If you are talking about Earth, there are some places where the acceleration due to gravity is different, but that is only due to abnormal events (Earth's buldge, mountains, etc.)

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AP®︎/College Physics 1

Course: AP®︎/College Physics 1 > Unit 4

Lesson 4: Newton's law of gravitation

Newton's law of gravitation review

Google Classroom

Review the key concepts, equations, and skills for Newton's law of gravity, including how to find the gravitational field strength.

Key terms

Term (symbol)	Meaning
Gravitational force ( $F_{g}$ ‍)	Attractive force between two objects with mass.
Gravitational field	A model explaining the influence an object extends to produce a force on other objects.
Gravitational field strength ( $g$ ‍)	The numerical value of the gravitational field at a point in space. SI units of $\frac{m}{s^{2}}$ ‍ or $\frac{N}{kg}$ ‍.
Inertial mass ( $m$ ‍)	Two objects have the same inertial mass if they experience the same acceleration given the same force. This is the same mass used in Newton’s second law. Experimentally equivalent to gravitational mass. Has SI units of $kg$ ‍.
Gravitational mass ( $m$ ‍)	The property of matter that causes it to experience a force in a gravitational field. Two objects that balance each other on a scale have the same gravitational mass. Experimentally equivalent to inertial mass. Has SI units of $kg$ ‍.

Equations

Equation	Symbols	Meaning in words
$F_{g} = \frac{G m_{1} m_{2}}{r^{2}}$ ‍	$F_{g}$ ‍ is gravitational force, $G$ ‍ is the gravitational constant, $m_{1}$ ‍ and $m_{2}$ ‍ are the point-like masses, and $r$ ‍ is the distance between the masses	The gravitational force between point-like masses $m_{1}$ ‍ and $m_{2}$ ‍ is directly proportional to their masses and inversely proportional to the square of the distance between them.
$g = \frac{F_{g}}{m_{2}} = \frac{G m_{1}}{r^{2}}$ ‍	$g$ ‍ is the gravitational field strength	The gravitational field strength is directly proportional to mass creating the field and inversely proportional to the square of the distance.

Newton's universal law of gravitation

Gravitational force

F_{g}

is always attractive, and it depends only on the masses involved and the distance between them. Every object in the universe attracts every other object with a force along a line joining them.

The equation for Newton’s law of gravitation is:

F_{g} = \frac{G m_{1} m_{2}}{r^{2}}

Where:

F_{g}

is the gravitational force between

m_{1}

and

m_{2}

G

is the gravitational constant equal to

6.67 \times 10^{- 11} \frac{m^{3}}{kg \cdot s^{2}}

and

m_{1}

and

m_{2}

are masses

The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers of mass. This is called an inverse-square law.

For example, if we double the distance between the Earth and the Moon, the attractive force between them would go down (because it is inverse), and it would go down by a factor of

4

instead of

2

(because of the square).

It describes both objects falling down and those in a circular orbit, such as a satellite around Earth.

How to find the gravitational field strength

All objects attract other objects by producing a gravitational field

g

, which is defined by the gravitational force per unit mass. We find the strength of this gravitational field of mass

m_{1}

on any object with mass

m_{2}

by dividing our above equation by mass

m_{2}

\begin{aligned} g & = \frac{F_{g}}{m_{2}} \\ = \frac{G m_{1} m_{2}}{r^{2} m_{2}} \\ = \frac{G m_{1}}{r^{2}} \end{aligned}

Common mistakes and misconceptions

Some people forget that gravity causes attraction between all objects. Every mass attracts every other mass. That means you are even gravitationally attracted to your friend, your pet, and even your pizza.
People forget that the force of gravity is inversely proportional to $r^{2}$ ‍ instead of just $r$ ‍. As the distance increases, the force of gravity decreases by a factor of $\frac{1}{r^{2}}$ ‍.
Sometimes people forget that $r$ ‍ is the distance between the centers of mass. We measure the distance between objects from their centers, not their surfaces.

Learn more

For deeper explanations of Newton's law of gravitation, see our videos:

To check your understanding and work toward mastering these concepts, check out the exercise on gravitational field strength and comparing gravitational and inertial mass.

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brandonmoe5529
Posted 3 years ago. Direct link to brandonmoe5529's post “why is finding the gravit...”
why is finding the gravity and accelerate Equations so confusing to me ?
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Lyn Kang
Posted 7 months ago. Direct link to Lyn Kang's post “Why do we distinguish bet...”
Why do we distinguish between inertial mass and gravitational mass?
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hallzhuu
Posted 3 years ago. Direct link to hallzhuu's post “Why does the equation for...”
Why does the equation for F_g not have a minus sign, since it's in the opposite direction of displacement?
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Iron Programming
Posted 3 years ago. Direct link to Iron Programming's post “Howdy, so I'm big on prov...”
Howdy, so I'm big on proving everything I learn so I was learning how to prove Newton's Law of Gravitation. I started to prove it using Kepler's 3rd Law and everything was going smoothly but then I saw somewhere that you used Newton's Law of Gravitation to prove Kepler's 3rd Law.

While both might be legit, it wouldn't be legit to think in circular reasoning: I must have an alternative way to prove at least one of them.

Newton's 3rd Law: F_g = G*(m1*m2)/(r^2)
Kepler's 3rd Law: P^2 = a^3*(4π^2)/(G(Msun + Mplanet))

So can you provide me with the appropriate background? I'm okay with other proofs that were created after Newton/Kepler but I don't want circular reasoning.

Also I would prefer the original proofs that Newton & Kepler used to prove their laws: unfortunately not near as many people teach proofs as concepts.
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- Carlos Castillo
  Posted 3 months ago. Direct link to Carlos Castillo's post “I would say that, to prov...”
  I would say that, to prove Newton's Law of Gravitation, you could most certainly perform an experiment, the Henry Cavendish Experiment.
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  (1 vote)
mehrin farjana
Posted 3 years ago. Direct link to mehrin farjana's post “When an object is thrown ...”
When an object is thrown straight up, its velocity decreases, and at its maximum height the velocity reduces to zero and then it falls back down. My question is, exactly how long does the object stays in the air at zero velocity, and does it (time) varies from place to place due to different gravity?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Inspiron13
  Posted 3 years ago. Direct link to Inspiron13's post “The object will stay at 0...”
  The object will stay at 0 velocity for an infintensimally small time period (it doesn't last long). To visualize this, let's say for example, the object reaches zero velocity 6 seconds when it is thrown. This means that at 5.9999999999999999 seconds, the object still has some velocity. Only at exactly 6 seconds the ball has 0 m/s. At 6.000000000000000000000000001 seconds, the object has velocity (which is really really really close to zero but not exactly zero). This means that only at that small point of time, exactly at 6 seconds (to infinite precision of digits), it will have exactly 0 m/s.
  
  To answer your other question, the time does vary from place to place due to different gravity. For instance, if you would try this same experiment on the Moon, it would take longer for it to fall back since the acceleration due to gravity is slower. (It would also go higher which would mean more distance).
  
  If you are talking about Earth, there are some places where the acceleration due to gravity is different, but that is only due to abnormal events (Earth's buldge, mountains, etc.)
  Comment on Inspiron13's post “The object will stay at 0...”
  (3 votes)
Najm
Posted 3 years ago. Direct link to Najm's post “Hey guys, Let's say we h...”
Hey guys,

Let's say we have a scale (not a balance) made on earth, this scale calculate the weight of an object and then provide it inertial mass, as it give the result in KG, so it must be mass not the weight.

Thus, this scale is probably using an equivalent of the formula of F=mg, while the g is the gravitational field constant on Eatrh.

Therefore, if this scale hasn't a feature to update the g value, it will provide different objects masses on other planets surfaces, and this would be wrong because the inertial mass is constant regardless of the object's location.

If my conclusion is right, the scales manufacturers must inform buyers that their products are only designed to work on Earth, in case someone would want to use them on another planet :) .

If not right, let me know why please!

Thanks
Najm
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(2 votes)
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noahreudink
Posted 3 years ago. Direct link to noahreudink's post “What is the acceleration ...”
What is the acceleration experienced by two things (55kg and 17000kg) that are floating out in space?
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dpena
Posted 3 years ago. Direct link to dpena's post “How do you find the mass ...”
How do you find the mass of a planet?
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- Anjali Gupta
  Posted 3 years ago. Direct link to Anjali Gupta's post “Can be determined using N...”
  Can be determined using Newton's law of gravitation.
  F = Gm1m2/r2 = ma (m1=mass of earth & m2=mass of object)
  Gm1/r2 = g
  m1 = gr2/G
  m1 = (9.8 m/s2)(6.37 × 106 m)2/(6.673 × 10−11 Nm2/kg2)
  m1 = 5.96 × 1024 kg
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  (1 vote)
dpena
Posted 3 years ago. Direct link to dpena's post “How do you find the mass ...”
How do you find the mass of the planet using gravitational forcE?
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(2 votes)
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ashikarikkalan002
Posted 4 years ago. Direct link to ashikarikkalan002's post “I just saw three equation...”
I just saw three equations of motion for a body thrown vertically downwards towards the Earth in a book which is given as v=u+gt,s=ut+1/2gt^2 and v^2=u^2+2gs. According to these equations, 1.the body is moving in a straight line 2.It has uniform acceleration 3. It's acceleration is equal to acceleration due to gravity
Which means it doesn't sounds like the body is under free fall
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(1 vote)
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- Ayush Rathore
  Posted 4 years ago. Direct link to Ayush Rathore's post “Hi there, You have a wron...”
  Hi there,
  You have a wrong intuition for these formulae. We have assumed the three points and then derived these formulae for an object to be freely falling under gravity 1.the body is moving in a straight line 2.It has uniform acceleration 3. Its acceleration is equal to the acceleration due to gravity. I hope you get it now 😊
  Comment on Ayush Rathore's post “Hi there, You have a wron...”
  (1 vote)