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### Course: High school physics—NGSS (DEPRECATED)>Unit 2

Lesson 1: Newton's law of universal gravitation

# Viewing g as the value of Earth's gravitational field near the surface

There are two ways to interpret the constant g = (-)9.81 m/s². The first interpretation is the acceleration due to gravity of an object in free fall near Earth's surface. The second interpretation is the average gravitational field at Earth's surface, which can be used to calculate the force of gravity on an object. Created by Sal Khan.

## Want to join the conversation?

• This video got me thinking. Suppose that my friend and I are tunneling downwards towards the Earth's center. As we proceed, I reason out that we should start to weigh less, since some of the Earth's mass that used to be beneath us is now "above" us (but not directly above us, of course) and is exerting a force that starts to contradict the force pulling us towards the Earth's center. Yet my friend reasons out that we should start to weigh more since according to Newton's law of universal gravitation, which states that F = G(m1)(m2)/r^2, the closer we get to Earth's center, the smaller the distance between the centers of our masses and of the Earth, which translates to a greater force.

We then think about what would happen to our weights when we reach the Earth's center. I say that we should become weightless, since there are equal amounts of mass pulling us in all directions, and therefore all forces should cancel out (we assumed that the Earth is a perfect sphere). My friend says, "No. Since the distance between the centers of our masses and of the Earth is zero, we would experience... infinite force?"

Which of us is correct? And what are your views on this?
• The law of gravitation treats the two objects as particles, meaning all the mass is located at one central point (this is Gauss's law, and only applies when the mass being influenced is outside of the sphere). However, in reality, the mass of the earth is distributed all around this blue sphere we call home. Although conceptually you are correct and your friend is wrong (forces of gravity influence you from all directions, ultimately canceling out), the pressure and heat at the center of the earth would compress/melt you to death before you could enjoy such a feeling of weightlessness.
• So, every time I trip over, I fall at an acceleration of 9.81 m/s^2 ?
• It actually depends upon exactly where on Earth you fall. g is not truly constant; it varies from location to location. 9.8 m/s^2 is just an average. The true value varies with your latitude and longitude (mainly due to surface variations in Earth's density). g also varies with altitude according to the formula: g/g' = (r'/r)^2, where g' equals g at the Earth's (or other celestial body's surface) and r' equals distance from the center of gravity of Earth (or other celestial body). For example, if you are at an altitude of 2.2x10^7 m (22000 km), the value of g drops to 0.512 m/s!
• What is the cause of gravitation?

Is it even known yet?
Does the theory of general relativity explain it or is it a fundamental interaction carried by a hypothetical graviton particle?
• From a quantum field theory perspective all forces are carried by particles referred to as gauge bosons, for gravity it would be the Graviton but we do not really have a quantum description of gravity that is mathematically consistent.

The best description we have of gravity is General Relativity and it depicts gravity as curvature of space-time.
• I don't understand what (kg*m)/s^2 means. How can you have kilogram meters? And how do seconds matter for the force experienced by an object on the ground?
• lets say we have a point mass(something as small as a point with some mass) now lets say u tie a string to it and pull it, to pull it you have to apply a force, and how would you define the amount of force you have you put in to flex your strength? like sal says pause and think, now sumin u have had a go at it, we work this out together, you will say, i pulled a mass of 'x' kilograms, 'x' meters away, in 'x^2' amount of time(aka the measure of the relative changes in your environment while you were at work). and thats the description of force. mass, distance, time, they are the dimensions, the descriptions we made to describe things in our real life. atleast thats my understanding of it.
• How were scientists able to determine that g is equal to 9.81 m/s^2? And how are they able to determine the force of gravity on other planets?
• By Newton's law of universal gravitation F1 = F2 = G* (m1*m2)/r^2
we multiply the Gravitational constant G = 6.673X10^-11 by the earth's mass divided by the earth's radius which will give us F/m2 = acceleration derived from the formula F = ma.
• At , it talks about the average gravitational field on the Earth's surface. I wanted to know how far does an object's gravity effect on matter? If it has no limit, even though it is negligible, does combing the distance and mass of objects near by give a combined for of gravity? (For example would the Earth's gravitational pull on the other planets be greater than 9.8 m/s^2, would it be the same, or based on the distance that other objects are in space it would decrease)

...and if theoretically there is no limit to how far matter's gravity is affecting other matter, than wouldn't that mean that the whole universe is pulling on itself?
• To answer your last question first, yes, there is no limit to the effect of gravity, and this does indeed mean that the whole universe is 'pulling' on itself. However, the strength of a gravitational pull gets weaker depending on the square of the distance between two objects, so at interstellar distances the gravitational effect of, say, our Sun would rapidly become very weak.

The Earth's pull at its surface is 9.8 m/s^2, but an object at its surface is only about 6400 km from the centre. The Moon is 384000 km away, which is 60 times as far, so the Earth's gravitational pull on the Moon is 60^2 (which is 3600) times as weak -- only 2.7 millimeters per second squared. Going out to other planets, Jupiter is 2000 times further away than the Moon, so the Earth's pull on Jupiter is a whopping 15 billion times smaller than the Earth's pull at its own surface. So the numbers do get very small very quickly.
• what exatly is normal force
• The acceleration in free fall is -9.81 m/s2 , but while skydiving you would be in equilibrium, but shouldn't it still be 9.8 m/s2 ?
• that's true because when you are sitting in a chair you are not free falling, and that chair is supporting your weight.
(1 vote)
• what would happen to g if earth stopped rotating
• The acceleration due to gravity is not related to the rotation of the earth. It is the gravitational acceleration caused by the mass of the earth.

If you are asking about the decrease in an objects measured weight because of the centrifugal force because of the rotation lets take a look at that.

The speed of a point on a circle is the radius times the angular velocity. The angular velocity of the earth's rotation is 0.00007272 radian/second the radius of the earth is 6,371,000 meters so you have 0.00007272 * 6,371,000 = 463 m/s at the equator. For centripetal acceleration you have a = (v^2)/radius so you have 463^2/6,371,000 = 214,369/6,371,000 = 0.034 m/s^2.

So for the value of g being 9.8 and the the centripetal acceleration being 0.034 which is about 0.34% of the value of g and depending on where you are on the earth the actual gravitational acceleration can vary by up to 0.7% so using 9.8 m/s^2 for problems like this is fine. If you end up doing actual engineering work where more precision is needed you will have to take the local gravitational acceleration and all other factors into your calculations.