Gravitation (4 of 17) Calculating Acceleration Due to Gravity (g)

Shows how to calculate the acceleration due to gravity. The equation is derived from Newton's second law and Newton's Law of universal gravitation.

The acceleration due to gravity is the acceleration on an object caused by the force of gravitation. All bodies accelerate in a gravitational field at the same rate relative to the center of their mass. A conventional standard value for this acceleration at the surface of the Earth is 9.81 m/s2. As you move away from the surface of the Earth the acceleration due to gravity decreases. The acceleration of the satellite is only dependent upon the radius of its orbit and the mass of the central object. The mass of the satellite has no effect on the acceleration..

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[thomasng]

THANK YOU SO MUCH I REALLY NEED TO UNDERSTAND THIS ESPECIALLY THIS TIME BEING STUCK AT HOME

[Exophite]

Thanks! my book made it nearly impossible to understand where the figures came from, its suprisingly hard to find a good video to explain things slowly and easily

[trumanhw]

Most people think that if you fall for 3 seconds ... you'll be traveling at
T-squared (3s squared = 9)
G (9.8 m•s ) ... which is NOT M/S (that says DIVIDE, not multiply) ...
or
9 • 9.8 m s^2 = 88.2 m•s ... right..?

Not at all. you're traveling at 29.6 m•s.

[MisterBinx]

9.81 m/s^2 is an approximate value we use for many calculations. It is actually different though depending on your actual distance from the center of the earth.

[royporterjr.]

Is it possible to rearrange g=9.8m m/s2 so s is not squared but instead m is squared? g=?.?s / m2?

[HawaiiSong]

How does it come out to Meters/Second^2 I am having trouble understanding how you get that 4:19 from the previous equation. Is m/s^2 supposed to be assumed, or does it cancel out somehow in the equation before hand.

[koutanakano]

4:15 Hi, could anybody tell me why km to m conversion turned 6, 361km to 6.371 x 10 to the 6th. I thought km to m only makes a 10 to the 3rd difference. Thanks.

[DeviantLife]

can you explain how what you did with all those units to arrive at meters per second squared?

[KB-svfm]

According to this, if a person fell at the top of a very tall building their speed of acceleration would be less than if they fell off a six foot ladder, right ? So at a greater height your rate of increase in speed would be less ? How would terminal velocity be affected, please ? Thank you.

[legionstark]

Thank you so much. This is so much easier than I thought it was going to be

[ayushroy]

Be careful, you wrote at 4:40 g = 9, 81 m/s^2 x 75kg = 736N..however g is not equal to 736N, you multiplied it by 75 to figure out the weight. Should have used ' = ' carefully :) Small error though!

[scottadams]

This was super helpful. Thank you so much!

[talgoam]

What happens when the radius approaches zero? Gravity approaches infinity? So the core of an appel has an infinitely high gravity?

[BAMBAMBAMBAMBAMval]

Is this the same equation used to calculate the gravitational pull of the center of the milky way?

[rubentuting]

now where does big G came from? why is it 6.67 × 10^-11

[silvereaglestudios]

The one thing I couldn't understand is what 6.67*10^-11 is supposed to mean.

[rossholst]

It is said that all objects will fall to earth at the same acceleration.

However this troubles me a bit. It would seem that per these equations, the acceleration of a dropped objects will be the same with respect to how quickly the objects are falling towards the earth.

However would these objects not also cause the earth to fall towards them?

If so, would the acceleration of the earth towards these objects not be dependent on the mass of the objects?

If instead we asked at what distance would the bowling ball and feather move the center of the earth 1m over the duration of the objects falling?

Now instead of dropping the objects close to the surface of the earth they were dropped from the distance where the bowling ball moves the earth 1 full meter. Would the objects still take the same amount of time before they reached the surface?

It seems that objects only fall with approximately the same rate. It being conditional on the objects being very small compared to the mass of earth and the objects being sufficiently close to earth.

If the objects are dropped from a greater distance the effect on how they also pull the earth becomes relevant. This requires a much greater drop height or a much more significant mass difference between the two objects (similar to the difference between the earth and the feather). If instead we compared dropping a feather and a planet…

It would seem that this result would be due to a unique scenario where essentially a single object accounts for nearly the entire mass of the 3 objects and where the distance covered is relatively small and not universally applicable when the mass is not dominated by a single object or the fall distance is very large.

[jospinlionel]

at 6:34 where did you find 122N? +Brian Swarthout

[anilsharma-evmy]

So 15 decimal value answer for g is possible
We can do calculates the next date of earthquake so we go to safe place ?Before it too late to calculate ?

[ohtych]

When we calculate the radius of the Earth, do you mean radius of M1 or half the distance TO M1?