Gravitational Fields | A-Level Physics | Doodle Science

[@PhysicsOnline]

Great summary of gravitational fields!

[@DoodleScience]

Thanks a lot!

[@nq2c]

Productivity it is less when you’re on top of a mountain( by a tiny amount). It is a constant on the surface of the earth, because everyone is at the same distance from the centre of the earth(pretty much)

[@dndjxnskdbajd4561]

@Productivity Gravitational Field Constant at Sea Level

[@DMFermiPhysics]

@Productivity Correct. The constant we use at the surface of the earth is generally linked to an elevation that is at sea level. The difference in the gravitational field at the top of Mount Everest is slight, only about 0.03 N/kg aka m/s^2 than at sea level. You can perform the calculation yourself while adding the height of Mount Everest to the radius of the earth though this does assume a uniform density for the earth but you'll get the point.

[@carultch]

@Productivity Not only does a high mountain reduce the gravitational field slightly, but also the fact that the Earth rotates, and the equatorial bulging of Earth, will also affect the apparent gravitational field at any given location. The strongest gravity would be at the north pole (9.83 N/kg), and the weakest apparent gravity (9.76 N/kg apparent gravity, 9.79 N/kg true gravity) would be at Mt Chimborazo in Ecuador. Since you can't feel true gravity, it is the apparent gravity that governs actions that take place on this planet. This is what you get after you "use up" some of gravity to cause a body to accelerate with the rotating Earth.

[@ratanake6261]

Who watch this video when ur test is tmr ?

[@PagManProductions]

@Akkash Eswaran good one

[@samkhalil3908]

My test is in 24 minutes

[@investzaku]

I have no test

[@AoiSkaia]

My test is in 10 mins

[@yung-sab6667]

Too bad my A Levels got cancelled smh

[@goldencrown6388]

Great video for a recap on this chapter! Loved it! Please keep making videos 😍😍

[@conspiracy11]

UR THE GOAT DOODLE SCIENCE I WISH I FOUND YOU EARLIER

[@harihartexochem8684]

Nancy I love this idea

[@arsalanhussainhaidery4244]

Sir found your content quite useful and easy... However it would be fascinating if you could post some more alevels physics sir.

[@josephwilson478]

We’re not in boot camp my friend.

[@Jonathan-cx9rc]

"Easy"??? Are y'all aliens or am I just too dumb?

[@Foresterscene]

@@josephwilson478 Sir! Yes Sir!

[@frederickthorpeyear7425]

@@Jonathan-cx9rc lmao too true

[@frederickthorpeyear7425]

@@josephwilson478 bros tryna join the army

[@syedumar6192]

Mind relax , help for quick revision , good resource....

[@venuraworld440]

I'm having my AL's today, hope I won't fail 😭

[@gauthampaul4213]

How did it go?

[@cindy1498]

Amazing revision material! Please upload some more!

[@amarsaeed9042]

Absolutely wonderful. Amazing summary.

[@maryamghayas6659]

I have a confusion, please tell at 1:46 why aren't you using the -ive sign?

[@mohammadmunzurulhaque3197]

there the concern is with the MAGNITUDE of the force particularly, which is always +ve anyways. The purpose of the -ve sign is, just, to show the force is always ATTRACTIVE... !

[@maryamghayas6659]

@@mohammadmunzurulhaque3197 thank you so much 😊

[@ontheway5792]

Such an amazing video! Thanks for ur effort....

[@joshuabiju1811]

arent the field lines supposed to be drawn in a way such that the arrow of the field line is in the middle of the field line

[@THEPRINCESSAMAZING]

This was amazing!

[@creative8d14]

Ok guys I’m gonna build a saucer and use a it-gravity and prank every one

[@xQryptic]

Excuse me, but where can I find the source for such a FIRE tune? asking for a friend, obvs :p

[@Yahye-ys1]

more A2 videos

[@eagleheart1634]

This feel very confusing for anyone else

[@piumisathsarani6044]

many thanks n keep uploading more of these

[@sahilm483]

very good summary

[@shayaanbawany7337]

THANK YOU!

[@TheLimitedPixel]

Can someone explain to me why we don’t use this in everyday examples, like how fast a parachutist falls?

[@carultch]

Because the region in which the problem takes place, is insignificant relative to the radius of the Earth. Consider Earth scaled down to the size of a 30 cm diameter classroom globe. The atmosphere would be as thin as the varnish applied to its surface, at this scale. The heights involved in when a parachutist falls from a plane are insignificant compared to the radius of Earth, that gravity might as well be uniform over the skydiver's journey. You could use the universal gravitation formula for a skydiver, but it would give you only a marginal increase in precision that will be significantly more difficult to calculate.

[@joshvictor110]

minute physics?

[@malakagunasekara3679]

so useful👍

[@josiebristow8192]

Awesome video thanks a lot!! also damn your writing is so satisfying

[@Music-ij1uu]

1:46 shouldnt it be 0.5 since the si unit for distance is in meters, and 50cm is 0.5m???

[@DoodleScience]

+George Washington Yeah that's right, but I just wrote 50x10^-2 m which is the same as 0.5m. It's just what I'm used to so I don't make silly mistakes with unit conversions :)

[@Music-ij1uu]

+DoodleScience oh okay i get it now, thanks and u earned a new subscriber :)

[@DoodleScience]

+George Washington Thanks a lot! :)

[@charliegodbold4183]

this helps me

[@notsurewhattoputhereyet7052]

ITS SO PERRRFEEEECT xD

[@sithumkarandawla286]

Why is F =-GMm/r^2

[@carultch]

#1, we know it has to be proportional to m, because this is consistent with experimental results. We know force of gravity is proportional to the mass of the object experiencing a gravitational force, since that is why scales can work as a proxy to indirectly measure mass. #2, because of Newton's third law, there needs to be symmetry to this formula. We need a commutative way to combine the two masses, to determine gravitational force on the original source of the gravity (M) from the original recipient of gravity (m). Switching source and recipient, makes m the source and M the recipient. Multiplication is commutative, so we can simply multiply them. This also satisfies part 1, when M is the recipient and m is the source. #3, because we live in a 3-D universe the sphere of influence of M's gravity is proportional to r^2, the distance between the two masses at their centers. The surface area of a sphere is proportional to its radius squared. As you get farther from M, M's gravity spreads out onto a larger sphere of influence. This means it will be inversely proportional to r^2. Had we lived in Flatland, there would be 1/r in this equation. Had we lived in 4D hyperspace, there would be 1/r^3 in this equation. #4, finally, in order to make it consistent with experimental results and practical to calculate in our system of units, we add a universal constant of proportionality out in front. This was first measured with the Cavendish experiment, and is now measured with modern reproductions of this experiment, if you are curious to look up how we know its value.

[@Ilovecruise]

Thank you

[@winfieldsheman5032]

nice !

[@imranzafar05]

hi Im also teaching physics in pakistan, can you tell me the name of software u used to make these videos? TIA

[@gagzs1660]

Subscribed

[@sahilm483]

thx a lot

[@notsurewhattoputhereyet7052]

For a uniform field, the equation is G=F/M... what about a field that's not uniform then??? :S I need to know this before the 19th but it probably won't happen :u

[@carultch]

The equation also applies for non-uniform fields, it's just that he had uniform fields on his mind at the time of recording that line. FYI, lowercase g is the field, big G is the universal constant.

[@adriangheorghe2327]

Constanta gravitationala G este un adimensional care la nivel macroscopic reflecta raportul intre suprafata integratoare a campului gravific si suprafata generatoare a campului gravific. Suprafata integratoare a campului gravific este suprafata care limiteaza corpul, suprafata pe care se aduna contributiile de camp ale tuturor particulelor din masa corpului. Suprafata generatoare de camp este data de suma tuturor sectiunilor generatoare de camp gravific, ale tuturor particulelor din componenta corpului. La nivelul nucleonilor constanta gravitationala Gn este data de raportul intre sectiunea de refulare a eterului si sectiunea de aspiratie a eterului. Raport egal cu 8.eps,0; Gn=8/(4.pi.k). Unde k este factorul interactiunilor electrice. Ceeace arata ca gravificul este derivat din electric. The gravitational constant G is a dimensionless that at the macroscopic level reflects the ratio between the integrating surface of the gravitational field and the generating surface of the gravitational field. The integrating surface of the gravitational field is the surface that limits the body, the surface on which the field contributions of all the particles in the mass of the body are collected. The field-generating surface is given by the sum of all the gravitational field-generating sections of all the particles in the body component. At the level of nucleons, the gravitational constant Gn is given by the ratio between the repulsion section of the ether and the aspiration section of the ether. Ratio equal to 8.eps,0; Gn=8/(4.pi.k). Where k is the factor of electrical interactions. Which shows that the gravitic is derived from the electric.

[@euan2312]

nice vid

[@camerondosanjh6040]

I agree

[@euan2312]

@@camerondosanjh6040 thanks for the feedback

[@leinyuymarie5570]

Great summary, but i cant help but notice that on the world map you drew, You didnt draw the African continent. I wonder why..

[@LoFortunato]

He did draw africa?!?!

[@Tokyo-ji2bi]

Watching this on the day of my p4 exam💀

[@roohiyactg3935]

Please make more videos

[@josephkaremera159]

Good

[@321silversnake]

Useful, would like to see more. But please make it a bit more original e.g. don't use the exact same music as minute physics

[@TheSecondJR]

I wach dis on seneca

[@unknown_person5905]

I didn’t understand , it is confusing for me

[@Rayrblx5]

Me too😂

[@motivationalbiography4726]

The equatiin is F=GMm/r2.... And not F=-GMm/r2

[@DoodleScience]

The minus sign is to show that the vector force is in the opposite direction to the direction of the radial unit vector.

[@lisagee3318]

Are they really really attractive I really didn't believe that but anyway

[@carultch]

So far, we haven't discovered repulsive gravity, so all gravitational forces that we know about are attractive. You can look up the thought experiment of negative mass, and you can see that even if negative mass existed, that it would still be attracted to positive mass, as long as gravitational mass and inertial mass are a package deal. You'd have to have negative gravitational mass and positive inertial mass, for repulsive gravity to exist. Either of these thought experiments, if they existed in reality, would be big problems for physics.

[@aonghuslo2998]

Bro you need way more explanation

[@rachael-personal6149]

[@tomdowning9006]

Shite

[@Yahye-ys1]

thank you