## Gravitational force and Gravitational constant

**gravitational force**, which you will be not knowing before,Here you will learn hidden facts of

**gravitational force**,I hope that you will enjoy this topic to know the hidden concepts of

**gravitational force.**Here in this topic your all doubt and confusion will be clear, After reading this complete post, I ensure you will feel that you have learned some hidden concepts about

**gravitational force**, which you were not learned before, so lets start to know gravitation force facts.

### Gravitational Force law history facts

As you know and learned by your teacher saying that an apple was fall on head of great scientist Newton, He did not ate that apple but started to think , Why did this apple fell down on the Earth ? And after that this gravitational force law was invented by Newton.

Ancient era before scientist Galileo, peoples believe was that Stars, Sun, Moon and other planet all moves around the Earth. Those time peoples were thinking that Earth is the centre of universe, But after so many hard works and experiment Galileo found that Earth is not the centre of universe, The Earth and other planets moves around the Sun. This was a great invention given by Galileo.

### Newton’s Law of Gravitation

From the above Formula F = Gm₁m₂/r² it is clear that if m₁ and m₂ will be heavy then force will be more. Similarly if r value will be large then force value will be small.

take an example if you living far distance from your family, Then your parents remember you less some time call you on mobile, but when you live with your parent at home then they call you all the times, means to say when distance is large attraction becomes less, but when distance is small then force of attraction is more ,Think how natural is this law and applicable for all point masses.

#### Gravitational force important points

**1 ** This gravitational force is a central force means this force will act along line joining their centre as shown in above picture F12(force on m1 due to m2) and F21(force on m2 due to m1), This force act at the centre of mass and act along centre line important concept for force direction.

Where r is distance from centre to centre for point masses like m₁ and m₂, G is universal gravitational constant and its value G = 6.67×10⁻¹¹ Nm²/kg²

**2** Gravitational force is always attractive force means two point masses attract each other in the universe. But you can think, why gravitational force is attractive in nature ? So the right answer for this question is the Almighty God made gravitational force attractive in nature to balance the Universe. If the gravitational force are not attractive in nature then planets, stars, Moon all were not doing motion and balances with each other Due to this attractive nature every things on the Earth are stable even you.

You must think the Earth is rotating then why not you displace from earth and fall down to other planet and in the universe.

Gravitational force attractive nature is the reason to stop everything fall down to other planet and in the universe. I hope you have got the importance of attractive nature of gravitational force.

**3** Gravitational force is independent of medium, means between two masses gravitation force measured in air, vacuum, water, oil at moon, you will find the value of gravitational force value will be same, Because G is universal gravitational constant, mass will not change and r will also not change so gravitation force value will not change.

**4** Gravitational force is same in magnitude but opposite in direction so it is written as

→ →

F12 = -F21 .

**5** Gravitational force is only valid for point masses like very very small masses not for big and heavy masses. Now this is important point to think over, which masses is big and which masses are small.

So for this concept you must think with respect to whole Universe, every masses in the universe are point masses. Now take two mass one Earth and other Sun these two masses are big and heavy but distance between these two masses are very very large, Hence these two are also look like point masses with respect to each other and treated as point masses, So gravitational force F = Gm₁m₂/r² is valid for Earth and Sun.

But Suppose this distance become small but two masses are big then this formula will not be valid you must remember this point.

**6** Significance of gravitational force is valid for heavy bodies, In day to day life you don’t experience learn many things from gravitational law.

for example two point masses attract each other, Then when you walk on the road and other people are also walking on the road. Hence according to this law you and other men get attracted and collide.

You are sitting in your room many masses are in room like TV, computer, fan, Wall all should attract and collide with each other but this does not happen Why ?

Because these all masses are small and Value of G = 6.67×10⁻¹¹ Nm²/kg² which is very very small, Hence this force value become very very small, which you not notice and feel .

But this force act, now in case of heavy masses gravitational force value become big and noticeable.

So gravitational force significance for heavy masses, but it does not mean you can not use for small masses, in the examination questions are based on small masses on gravitational force because examiner just want to know your concept of gravitational force.

#### Gravitational force questions

From above first question Q1 .

You know that F = Gm₁m₂/r² , how you will take step to solve this question, here both mass is identical given from question, but m is not given, so you have to suppose mass m for both the masses because both are identical and made by same material.

Now you see both the mass are spherical, suppose density for both masses ? hence you can now find mass m = volumexdensity = 4??³/3 x ? since volume of sphere =4??³/3

Now apply gravitational force formula F = Gm₁m₂/r² = Gm²/(2r)² here m1 = m2 = m and r = 2r distance between centre to centre now put value of m in above equation of force you will get F = Gm²/(2r)² = G (4??³/3 x ?)²/4r² here all G , ?, ? are constant after simplification F will be proportional to **r⁴** so (c) is correct answer.

Now from question Q2 you have to check which statement given from a to h are correct .

→ → → →

(a) F12 = – F21 F12 = GmM/r²i^ and F21 = –GmM/r²i^ you have already seen this statement is correct, because forces are same in magnitude but opposite in direction.

→ →

(b) Fnet = 0 now you see from from expression F12 +F21 = 0 this means that total external force on the system is zero F12 and F21 are internal forces so this is also correct statement.

→ →

(c) a₁₂ = a₂₁ for this calculate a₁₂ = F/m = GmM/(mr²) = GM/r² so a₁₂ = GM/r² now similarly calculate acceleration a₂₁ = F/M = GmM(Mr²) = Gm/r² so a₂₁ = Gm/r² hence clearly seen that a₁₂ value is not equal to a₂₁, you can also feel one mass m is small and mass M is big, force on both mass is same which one will move faster obviously small mass will move faster so its acceleration will be more than acceleration of M so this is wrong statement.

→

(d) acₒₘ = 0 that is acceleration of centre of mass, Since Fnet on the system is zero so acceleration of centre of mass must be zero. where as both the mass have different acceleration but centre of mass acceleration will be zero how and why ? this is beauty of Physics .You have already studies in previous post for reference you can refer centre of mass Why answer is very simple.

You have already studies when Fnet = 0 means net force on any system is zero then acceleration of centre of mass will be zero.

How for this you have already studies in previous post if you not please refer centre of mass post, where centre of mass formula acₒₘ = (m₁a₁+m₂a₂)/(m₁+m₂) now you have already calculate a₁ and a₂ value above and here m₁ = m and m₂ = M put the value in equation

acₒₘ = (m₁a₁+m₂a₂)/(m₁+m₂) , you will get acₒₘ = 0 . So you can understand by any point either by why or by how but result will be same.

Hence this statement is correct .

→ →

(e) V1 = V2 = 0 this is clearly seen that both the masses have different acceleration a₁₂ and a₂₁ so their individual velocity will not be zero, So this statement is wrong without any doubt.

(f) Vcom = 0 Since acₒₘ = 0 so velocity of centre of mass Vcom will be constant it will not be zero. hence this statement is wrong.

(g) Vcom = constant , This statement is correct .

(h) V1 = V2 = constant , This statement is wrong because both the masses have different acceleration a₁₂ and a₂₁ so their individual velocity will not be constant.

But centre of mass velocity will be constant.

### Does the Earth accelerate towards you ?

Does the Earth accelerate towards you, when you jump from a height ? You may find it hard to digest but the correct answer is Yes. You don’t see and feel the acceleration since it is very small, but you do move the Earth just like the Earth moves you due to gravity, let me explain see the below picture.

There are equal and opposite forces on you and the Earth due to Gravitational attraction |

Gravitational force acts between all objects with mass. The simple facts that an object has mass means that will attract with other object with mass. As you have seen the gravitational force magnitude between the objects depends upon their masses.

#### Earth accelerate towards you

The mass of the Earth is approximately 6,000,000,000,000,000,000,000,000 kg. that is 6×10²⁴ kg , Now the average human has a mass of 62 kg. So the mass of the Earth is relative to you is very very very large.

Now you know from Newton’s second law, for a given forces F = ma on an object the acceleration reduces as the mass of the object increases.

Hence it is easy to see that the acceleration of the Earth compared to you, When you are falling is incredibly small. However the Earth still accelerates towards you even if it is very small.

#### Gravitational force application

Gravitational force application |

From the above picture two body of mass 10kg and 5kg are initially at infinite distance and both are at rest. After releasing both the masses start to move towards each other due to gravitational force of attraction.

At any time it is found that the velocity of 10kg mass is 4m/s, then what will be velocity of mass 5kg ?

How you will solve this types of question, apply the concept there are two state one is initial and other is final. Now see the net external force in both the cases is zero, You must remember that gravitational force is internal forces between two masses not external force.

Hence Fnet = 0 so acceleration of centre of mass acom =0, Then velocity of centre of mass Vcom = constant. **Big concept **Vcom is constant means if it is initially zero then finally it will be also zero. So find out initial Vcom = (m₁v₁ + m₂v₂)/(m₁ + m₂) but as you see initially both the masses are at rest so v₁ = v₂ = 0 , Hence Vcom = 0 (Initially).

This means that final Vcom will be also zero because Vcom must be constant, So now find final Vcom = (m₁v₁ + m₂v₂)/(m₁ +m₂) for final case v₁ = 4m/s and v₂ =v and m₁ = 10kg and m₂ = 5kg also Vcom = 0, now put in above equation you will get.

0 = (10*4 + 5*v)/(10+5) or v = -40/5 = -8 so v = -8m/s .

Understand Concept individual both masses will have acceleration and velocity but acceleration of the system will be zero and velocity of the system will also be zero as Vcom is zero.

### Superposition of Gravitational forces

Superposition of gravitational forces means, You know that among every two point masses gravitational forces acts, but think if there are more than two masses then what will happen ? how gravitational forces will acts on the system of point masses.

This is done by superposition principle. According to superposition principle, you take a single mass and rest all masses of the system will attract that single mass, Add all the gravitational forces by vector addition rule on a single mass due to other point masses.

This is superposition principle, see the below question.

From the above picture, you have to find the net forces on mass m and 2m. From the question it is given triangle is equilateral having side length a. So each angle of triangle is 60º.

Now first calculate net force on 2m, as shown in the figure on 2m two force are acting due to two masses m. So the net force on 2m is F and F angle between them is 60º., so now you can calculate by using vector resultant formula.

Fnet = √(F² +F² +2*F*Fcos60º) = F√3, Hence net force on 2m will be √3F

Now you know for gravitational force F = Gm*2m/a², So Fnet = 2√3Gm²/a².

Similarly net force on mass m, here also two forces acting on m, F1 and F so F1 = Gm*m/ a². or Gm² / a²., F = Gm*2m/ a² = 2F1 .

Hence again use the resultant formula Fnet = √(F1² +(2F1)² +2*F1*2F1cos60º) = √7F1

Net force on m = √7*Gm²/a². So in this way you can find out net force on any mass by using superposition principle.

### Why Gravitational force is applicable only for point masses ?

According to Newton gravitational force formula is only applicable for point mass, Everywhere i have mentioned point mass.

Now this is time to understand feel of point mass important concept about gravitational force, You will learn the best concept for point mass here.

See the below question to explain the concept of point mass for gravitational force.

Gravitational force acts on point mass |

From above picture, How you will try to solve gravitation force between rod of length L and mass M and a point mass m.

Think rod is not a point mass its mass is distributed over the length L so what you will do to solve, The whole mass of the rod will be at centre of the rod, So you can suppose the whole mass of the rod at centre M, Now the distance between M and small m will be (L/2 +a) because length of the rod is L.

So gravitational force between rod and small m will be = Fg = GMm/(L/2+a)²

Hence gravitation force Fg = GMm/(L/2+a)² if you have done this then you are **WRONG**

now you will ask why this is wrong ?

Applying centre of mass concept is wrong here because every mass of the rod is not attracting the small mass m with equal force.

Remember gravitational force formula F = GmM/x² inversely proportional to x this means that the distance x is less force of attraction is more.

See from the picture mass divided into small dm, dm₁, dm₂ …… If all this mass get attracted with same force then your logic of centre of mass was correct but you see here all mass will attract with different force because distance x will be different for different dm,dm1,dm2,dm3………. dm will be more attracted than dm1, dm2 …. and so on .

So some masses near small m are attracting more and some masses are far small m will attract less. Hence rod one side more attraction and other side less attraction so your logic for centre of mass is wrong here because x is inversely proportional.

But your logic was correct in case if the distance was directly proportional for example in case of potential energy U = mgh¹ here h power 1 . Hence here your logic will be correct .

So how you will solve this question, applying the concept of small mass dm so Newton explained that gravitational force formula is only valid for point masses.

Now i hope your concept for point mass is clear.

So from the picture a dm mass taken at a distance of x hence gravitational force on dm due to m mass will be Fm = Gmdm/x² , now this formula is valid because dm is very vey small mass like point mass.

So divide M of length L in many small dm masses, Now according to superposition principle if there are many point masses then net force will be vector some of all the forces.

Hence Fnet will be like this.

→ → → →

Fnet = Fdm₁ +Fdm2 +Fdm3 +……………….

All small dm will attract mass m in the same direction so simply add all forces removing vector symbol Fnet = Fdm₁ +Fdm2 +Fdm3 +……………….

Now you know how to add this types of small quantity tell me, I hope your answer will be by integration so use integration.

**∫dF =∫ Gmdm/x² ** now check limit first dm start at distance a and last dm end at distance (L+a) so limit will be from ‘a’ to (L+a) hence .

L+a L+a

**∫dF =∫ Gmdm/x² = Gm∫dm/x² ** now there is a problem you can not integrate dm with x² so

a a

for this you have to change dm in term of x, use unit mass concept for L length mass is M so for dx length will be dm = M/L*dx so put this value in above equation .

L+a L+a

Fnet = **Gm∫dm/x²** = Gm∫M*dx/L*x²

a L+a a L+a

Fnet = GmM/L ∫ dx/x² = GMm/L⎜-1/x⎥

a a

After putting value of limit you will get **Fnet = GmM/(a(L+a)) **

Now for this moment i hope you have enjoyed learning gravitational force and gravitational constant. I will continue Gravitation post with more concepts and questions wait.

I also want your feedback through comment like and shares Thanks.

Dated 3rd Dec 2018

## 2 Comments

## Rajwinder Khatra · February 27, 2019 at 10:40 am

Concepts are very nicely explained from very basics to in depth . Very interesting reading . My congratulations to the author for developing such a fine website on the laws of Physics . Good to read .

## admin123 · February 28, 2019 at 6:43 pm

Thanks for your comment