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1 COMPOSSED AND WRITTEN BY PROF. NAJEEB MUGHAL. GOVT. MUSLIM SCIENCE DEGREE COLLEGE HYD.

1

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CHAPTER 6 gravitation

Contents:

1.

NEWTON’S UNIVERSAL GRAVITATIONAL LAWS

2.

CALCULATION OF MASS AND DENSITY OF EARTH

3.

VARIATION OF “ g ” DUE TO HEIGHT AND DEPTH

4.

5.

6.

WEIGHTLESS NESTS IN SATELLITES

ARTIFICIAL GRAVITY

EQUATIONS

7.

DIMENSIONS

8.

SHORT QUEASTIONS AND ANSWERS

Gravitation:



Gravitation is a natural phenomenon by which objects with mass attract one another. It is responsible for keeping the

Earth and the other planets in their orbits around the Sun; for keeping the Moon in its orbit around the Earth, for the formation of tides; for convection (by which hot fluids rise); for heating the interiors of forming stars and planets to very high temperatures; and for various other phenomena that we observe. The simpler Newton's law of universal gravitation

provides an approximation for most calculations. The terms gravitation and gravity are mostly interchangeable in everyday use, but a distinction may be made in scientific usage. "Gravitation” is a general term describing the phenomenon by which bodies with mass are attracted to one another, while “gravity" refers specifically to the gravitational force exerted by the

Earth on objects in its vicinity. This was studied by a scientist named Sir Isaac Newton in 1666.

DESCRIPTIVE PART

1.



Newton’s’ Universal Gravitational Law:



 In the universe every point mass attracts every other point mass by a force pointing along the line joining their centers. That two bodies attract each other with a force proportional to their product masses and inversely proportional to the square of distance between them.

Explanation:

Consider two masses m

1

and m

2

, if any one mass increases then product increases and any one mass decreases then product decreases. According to the first part of the statement,

Thus,

F



m m

2

The separation of masses is “r” distance apart. If the distance between the masses is made half, the force increases four ti mes or distance made double the force become one- four, according to the second part of the statement.

F



1

2 r

Mathematical Derivation:

F

 m m

1 2 r 2

This is mathematical form of Newto n’s gravitational law.

F = G r

2

2

Vector form:

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2




The direction of force is represented by unit vector. And the negative sign indicates the force of attraction.

Hence,



F = - G m m

1 2 r

2



 u

Applications

There are numerous applications of Newton's Universal Gravity Equation.

Earth and Moon

What is the gravitational force felt on the Moon from the Earth?

The Earth and Moon are 3.84x10

5 kilometers apart. Since the units of G is in meters, you need to change them to 3.84x10

8 m.

Note that the distance is from center to center. The force on the surface of the Earth would be slightly less, but since the distance to the Moon is so much greater than the radius of the Earth, the difference would be negligible.

The mass of the Earth is 5.98*10 24 kg and that of the Moon is 7.35x10

22 kg. The force of attraction on the Earth from the Moon is:

F = GMm/R² = (6.67x10

-11 Nm²/kg²)(5.98

x10 24 kg)(7.35x10

22 kg)/(3.84x10

8 m)² = 1.99

x10 20 N

This considerable force is what holds the Moon in orbit around the Earth.

Effect of Moon on person

The gravitational pull from the Moon on the 50 kg (110 pound) girl is:

F = GMm/R² = (6.67x10

-11 Nm²/kg²)(50 kg)(7.35

x10 22 kg)/(3.84x10

8 m)² = 1.67

x10 -3 N = 0.00167 N

She would not notice the pull from the Moon, since the gravitation pull on her toward the Earth is 490 N. But still, she is attracted more toward the moon than the boy who was sitting next to her.

2. Calculation of mass of the earth:


Newton said every body attracts every other body with a force, that bodies is,

F = G m M e .

R

2 e

Newton also said that the acceleration of an object is directly proportional to the net force acting on the object, is in the direction of the net force, and is inversely proportional to the mass of the object. F = m g

By comparing above two equations, m g = G m M e

R

2 e g = G

M e

R

2 e

We know that, g=9.8m/sec 2 , R e

= 6.38 x10 6 m,G=6.673x10

-11 Nm 2 /Kg 2 and calculated value of mass of the earth

M e

3. Calculation of density of the Earth:


=5.98x10

24 Kg

We know that density is the mass per unit volume, denoted by, “  ". The density of the earthy is given by, ρ =

M e

V

The volume of the earth is,

V =

4

π R

3

3 e

Therefore, ρ =

M e ρ =

3

4







M

π R e

3 e







4 π R

3

3 e

R e

= 6.38 x10 6 m

M e

= 5.98x 10 24 Kg and the calculated value of density of the earth,  = 5.52x10

3 Kg.m

-3

4. Variation of “g” due to altitude:








3





The acceleration due to gravity g is a variable quantity and it varies from place to place. We here, will derive the relation that describes variation of g with altitude. Let us consider a body of mass “m” lying on the surface of earth of mass and radius R e radius of earth. Let g be value of acceleration due to gravity on the surface of earth. We know that g = G

M e - - - - - - - -  (1)

R

2 e

Suppose the body is taken to height ‘h' above the surface of earth where the value of acceleration due to gravity is g h

.

Then g = G h

 e

M e



2

- - - - - - - -  (2)

Divide equation (2) by eq. (1) we get, g h g

=

G





 e

G



M e



 R

2 e





2



 g h g

=

2 R e



R + h e



2 g h g

=



R 2 e

R 2 e



2 by dividing with

R

2 e

R

2 e g h g g h g

1

=







1 +

= 1 +

R e h h

R e





2







2 g h g



= 1 - 2

 h

R e

     







From binomial theorem

We get g = g 1 - 2 h



 h

R e





thus, acceleration due to gravity decreases with increase in height.

5.

Variation of “g” due to depth:


Suppose “m” is the mass of an object placed on to the earth’s surface having “R e

” radius.








4




We know that g = G e

M e

R

2 e

- - - - - - - -   1 

Where, ρ =

3

4







M

π R e

3 e







Or

M = ρ π R e

4

3

3 e

The equation  1  can be written as g e

=

G





4

3

ρ π R

3 e





R

2 e

Or g e

= G





4

3

ρ π R e





- - - - - - - -   2 

When object into be taken at depth “d”, in side the earth, g = G d



M e

R - d e



2

- - - - - - - -  3 

Where e

4

M = ρ π R - d

3

 e



3

The equation  3  can be written as, g = d

G





4

3

 

3



R - d e



2





Or g = G d





4

3

 e







- - - - - - - -  4 

Divvied eq  2  by  4,  we get, g d g e

=

G





4

3

 









4

G ρ π R

3 g d







R - d e



= g R e e g = g d e



 1 - d

R e





This shows that, if R e

= d Then, g d

=g  1 – 1 

g d

= 0 It means that value of “g” is zero at the center of earth.

6. Weightlessness in Satellites:

In an accelerating system, the apparent weight of an object is not equal to its true weight. If the system is free falling, all supported forces must be equal to zero and object appear to be weightless.

Explanation:

 When a man is in such a satellite, he will fell himself weightless. This state is called “Weightlessness.”








5




To explain the weightless in satellites, suppose an object of weight ‘w’ is suspended in the roof of an elevator by means spring balance:

Case 1: Elevator at rest

Suppose an eleva tor is in the state of rest, Newton’s second law of motion, acceleration is equal to zero, the resultant forces on the object are equal to zero

F = m a

F' - m g = O

Or

F' = m g

Case 2: Elevator with constant velocity

If the velocity is constant, then the acceleration is zero. The apparent weight is equal to actual weight.

Case 3: Elevator accelerating upward

Suppose an elevator is moving upward with acceleration ‘a’, along positive y - axis. According to second law, the unbalanced force



F' - m g



acting on it.

Therefore

 

 

The apparent weight of an object is more than its rest weight.

Case 4: Elevator accelerating downward

Suppose an elevator is moving downward with acceleration ‘a’, along negative y -axis. According to the second law of motion, the unbalanced force

 m g - F'

 acting on it,

Therefore

Or



 

The apparent weight of an object is less than its rest weight

When a = g

Then

F' = m g - m g

F' = 0

And the object appears to have no weight; it appears to be “weightlessness”.

7. Artificial gravity:

When the astronauts are their spacecraft they fell inconvenient in their chambers. To solve this problem an artificial gravity can be created in the spacecraft by spinning it, about the axis

Suppose a spacecraft has two chambers by a tunnel of length 20 meters. Let us calculate the frequency to provide artificial gravity

We know that, a = c

2

4 π R

T

2 a = 4 π R c

2



2

Where “R” is the half length of tunnel,

a c

2

4 π R

2

= ν

a c

2

4 π R

= ν








6




If a body falls freely, then a c

= g Therefore,

1 g

2 π

R

= ν

1 9.8 m /sec

2

2 π

10 m

= ν

 = 0.158 rev. / sec

Hence, astronaut feel easy in space craft under 9.6 rev. / min at a distance of 10 meter away from point of axis of rotation

Equations

1.

F = G m m

1 2 r

2

2.

g R e

2

G

= M e

3.

g = g 1 - 2 h



 h

R e





4.

g = g d e



 1 - d

R e





5.

g d g e

=



R - d e



R e

6.

g h g

7. ρ =

3

4







M

π R e

3 e







8.

1 g

2 π

R

= ν 9. g = G

M e

R

2 e

10. g = G h 

M e



2

11.

g = G d

M e

 

2

Dimensions

=

2

R e



R + h e



2

12.

v =

2 G M

R

PHYSICAL QUANTITY & SYMBOL DIMENSION UNIT

Gravitational constant G =

F r

2 m

2

-2

M L T L

2



M

2 

-1 3

M L T

-2

N.m.

2 kg

-2

.

Mass of earth

g R 2 e

G

= M e



LT

-2

L

2

-1 3

M L T

-2

kg

Density of earth ρ =

3

4







M

π R e

3 e







L

3

= M L -3 kg.m

-3 .

Frequency

1 g

2 π

R

= ν



LT

-2 

1

2

  1

2

Acceleration due to gravity g = G

M e

R

2 e

= T

-1



sec -1 or hertz

M L T -2

 

L

2



-2



m.sec

-2

.

Short questions

Q.# 1 Is there no gravity in space?

Answer:

No it is not correct. If there were no gravity in space, the space shuttle would not be able to orbit the Earth, the moon would not orbit the Earth, and the Earth would not orbit the Sun. The reason we tend to think of there being no gravity in space is that we have seen movies of the astronauts being "weightless". They aren't actually weightless, they are still being pulled down by gravity but they and the space shuttle are in a constant state of freefall around the Earth. So they seem to be weightless as a result of the falling - just as you would seem weightless if you were in an elevator when the cable broke.








7




Q. # 2 Are “G” and “g” are the same thing.

Answer:

No, it is not correct. Where “G” is the universal gravitational constant and “g” is the acceleration due to the force of gravity and its value of 9.8m/s/s down is only true on this planet.

It is not a universal constant.

Q.

# 3 How would the weight of the body vary when it is taken to the center of the earth?

Answer:

The weight of the body becomes zero at the center of the earth because the acceleration due to gravity is zero at the center of the earth.

Q.

# 4 How the weight of a body vary as it is taken from the earth to moon? What will be the effect on its mass?

Answer:

As the force of gravity is inversly proportional to the square the distance, the weight of the body will decreases. As the body at the surface of moon is found to be

1

6 g = m

  g e

.The mass of a body will be unaffected. th on the weight on the surface of earth.

Q. #5 How the value of gravitational force be affected if the distance between two masses becomes halved?

Answer:

We know that, F = G m m

1 2 r

2

F' = G m m

1 2 r

 

2

F' = G

4 m m

1 2 r

2

F' = 4 F , this shows that the gravitational force increases four times, if the distance between two masses becomes halved.

Q. #6 Is the gravitational force between the earth and sun is same all the time for year?

Answer:

No, the gravitational force between sun and earth is not same all the time of a year.

Q. #7 How the weightlessness is to be controlled in satellites?

Answer:

The weightlessness in satellites is to be controlled by producing artificial acceleration due to gravity.

Q.

# 8 How the gravitational force would be affected if the each mass become double.

Answer:








8




F = G 2 F' = G r

2

2 F' = G r

2

2 F' = 4F , This r

2 shows that the gravitational force increases four times, if the each mass become double.

Q. #9 Comment on "g" variation due to height and depth.

Answer:

The value of "g" decreases by



2h



 R e



times as height increases. Because, g = g 1 - h





 towards the center of earth at depth by

 d



 R e

 time. Because g = g 1 - d



 d

R e





.

2h

R e







.And again it decrease

Q. # 10 If any one goes away from the earth surface equal to the earth radius above the earth surface. The value of "g" will become i) one fourth ii) one eight iii) one ninth.

Answer:

If any one goes away from the earth surface equal to the earth radius above the earth surface. The value of "g" will become decreases by one fourth.

Q. # 11 What will be value of "g" at the center of the earth?

Answer: The value of "g" the center of earth becomes zero. . Because, R e

= d g = g 1 - d



 d

R e





And, g d



0

Q.

# 12 State Newton's universal gravitational Law.

Answer:

The magnitude of force of attraction in the universe between two masses is directly proportional and inversely proportional to the square of the distance between their centers. It is known as Newton's universal gravitational law.

F

 m m

1 2 r

2

Q. #13 If the mass of the earth becomes nine times and its radius three times. What will be the value of g?

Answer:

We know that, g =

G M e

R

2 e

= 9.8m/sec

2

.

g' =

G 9 M e

2 g' =

G 9 M e

9 R g' =

G M

  e e g' = g , this shows that no effect on to acceleration due to gravity, if such change takes place.

Q. # 14 Is “g” is gravity.

Answer:

It is not correct that “g” is the effect of the force of gravity. Gravity is a force, little “g” is an acceleration caused by gravity.








9




Q. # 15 How would the gravitational force be affected if the one mass become double and the distance between them also double?

Answer:

We know that, F = G 2 r

2

F' = G

2 m m

1

 

2

2

F' = G

2 m m

1 2

2

or F' = G

m m

1

 

2

2

1

F' = F

2

, this shows that force become halved








10




Calculation of mass of sun:


We know that earth is orbiting around the sun. The distance between sun and earth (R) is 1.49 x10 11 m, G = 6.673x10

-11 N m 2 kg -2 And T = 365.25 x 24 x60 x 60 sec

We know that,

F =

G M m s

R

2

- - - - - - - -  ( 1)

F = m a

F = m





2

4 π R

T

2





- - - - - - - -  (2)

By comparing eq (1) and (2) m





2

4 π R

T

2





=

G M m s

R

2





2 3

4 π R

G T

2





= M s

2 11 3

4 (3.14) (1.49×10 )

M = s

2

1.304 × 10

35

6.64 × 10

4

30

M = 1.96 × 10 Kg s

Question (36): Name the different layers of the earth?

Answer: The core, mantle and the crust.

What is the acceleration due to gravity on the moon?

In: Physics [ Edit categories ]

[ Edit ]

[ Edit ]








11




The acceleration due to gravity can be found using Newtons law of Universal Gravitation.

The mass of the moon is 7.36x10^22 The radius of the moon is 1,737.4 kilometers or

1.7374x10^6 meters and newtons gravitational constant is 6.67x10^-11(m^3)/(kg*s^2).

Using this information we can use the formula a=(G(m2))/(r^2) where a= acceleration on the moon G is gravitational constant and m2 is the mass of the moon.

So... a=(6.67x10^-11(m^3)/(kg*s^2)(7.35x10^22kg))/(1.73x10^6m)^2 so acceleration on the moon a=1.62631m/s^2

So what is really mass and weight if they are not the same thing?

Mass is defined as the amount of matter an object has. One of the qualities of mass is that it has inertia As an example of inertia, imagine an ice puck resting on a frozen pond. It takes a certain amount of force to set the puck in motion. The greater the mass the more force will be needed to move the puck. The same is true if the puck were sliding along the ice. It would continue to slide until a force is applied to stop the puck. The more massive the puck is, the more force will be needed to stop the motion of the puck. Mass is a measure of how much inertia an object shows.

The weight of an object on earth depends on the force of attraction (gravity) between the object object and earth.

We can express that force as an equation:

F = G[M m/r 2 ] , where F is the force of attraction, M is the mass of the earth, m is the mass of the object, and r is the distance between the center of mass of the two objects (G is called the Gravitational Constant)

What does this equation show? What will cause the force of attraction to increase or decrease? If either mass increases the force of attraction increases proportionally. Since the moon has 1/6 the mass of earth, it would exert a force on an object that is 1/6 that on earth.

Why is the 1/r 2 factor so important? This is an inverse square relationship which seems to show up a lot in physics.

How does it affect the force?

What is 1/r 2 when r=1, 2, 5, 10? What is the decimal equivalent? Notice that when r=1 the value 1/r 2 is 1.0, but at r=10 it deceases to 1/100. That means gravity gets weak 'quick' as we move away from the earth.

To get a real feel for the inverse square relationship, see if you can get two magnets. Move the poles closer and closer slowly, what do you notice when r (the distance between the poles) is very small?

 How long for gravity to pull objects together?

 What does the F stand for?








12




 How does Einstein's theory apply to Earth?





Would a B-B and bowling ball fall at the same rate?

 Gravity and a spaceship

Weight inside and outside the body

 Where does gravity come from?

Next 10

How long for gravity to pull objects together?

Question

January 27, 2009 dear teacher

Two bodies, of M and m kg respectively, are far from each other about r meters. They begin to move one toword the other according to the law of gravitation. How long (Seconds) would it take until they meet?

Can you show, briefly, how you get the result? jacob - Israel

17162

Answer

Unfortunately, that is a very difficult problem to solve mathematically, requiring Calculus. With an object falling to the Earth, the acceleration is approximately constant, so you can easily calculate the time it takes. But with two objects attracting to each other, the acceleration increases the closer they get.

Sorry, but we don't have that derivation available.

Back to top

What does the F stand for?

Question

January 19, 2009

I am The beginner. What does the F stand for?

- USA

17115








13




Answer

In the Universal Equation paragraph, F is defined as the force of attraction between two objects, due to gravity.

This is a tough lesson, because it has more math. It is good to check the other gravity lessons to get a good background.

Back to top

How does Einstein's theory apply to Earth?

Question

December 3, 2008 your article about gravity is very helpful and thank you but help me learn some more. how does einstein general theory of relativity apply to us on earth? i have trouble applying the often used schematic drawing of the warped spacetime fabric on me and the earth as it is used(one big mass indenting the fabric and orbited by a smaller mass or planet also doing its smaller indentation on the fabric). why would the smaller orbiting mass not eventually crash into the bigger mass as the orbital path gets closer? if gravity per einstein is no longer a real pulling "force", why is it still considered one of the four fundamental "forces" of nature. If i can fall and pass through the earth like a neutrino, will i stop falling when i get to its center?

- USA

16893

Answer

Consider Newton's law of inertia that a moving object travels in a straight line. Since Einstein says that matter distorts space, an object traveling along a "line" in space will follow that line and fall toward the mass. That is a clever way to explain the gravitational force, but modern views state that gravity is caused by an exchange of graviton particles or that it is a result of a force field of some sort.

You could draw a 2-D picture of the lines in space curving toward an object, with the curvature getting greater for closer lines. Trying to draw or visualize it in 3-dimensions is difficult.

Note that a neutrino could be affected by a gravitational field, just as a photon is. But the neutrino also can easily pass through huge quantities of matter before it collides with an atom.

Back to top








14




Would a B-B and bowling ball fall at the same rate?

Question

November 25, 2008

I assume the Force created by gravity is based on the mass in question. However, or illustration purposes, would it be safe to say that a B-B and a Bowling Ball fall to the earth at a rate that is essentially the same? 9.8 m/s2?

Thanks. stan - USA

16853

Answer

For objects relatively close to Earth, the equation F = mg holds and all object fall at the same rate, independent of the mass of the object PROVIDED the effect of air resistance is negligible. A B-B and bowling ball would fall at the same rate in a vacuum, but in the atmosphere the air resistance would slow down the B-B.

Back to top

Gravity and a spaceship

Question

October 9, 2007

I'm having a little problem understanding gravity. I am trying to recreate space via programming. How am I suppose to translate a newton or F value into something i can understand or use? Ex: Your flying through space in a ship(imaginary). How close to u need to get to a planet before u can feel it ? Given the ships mass and all the necessary info. I am looking for something like 983cm*sec*sec(even having problems understand the sec*sec) with its maximum range of the gravity if any. If I am way off please let me know. This is for pure curiosity.

Thank you

Peter - Canada

14501

Answer

If you had a spaceship of mass "m" and you approached a planet of mass "M" at a distance "R", then the force pulling your spaceship to the planet would be F = G*M*m/(R*R), where G is the universal gravitational constant.








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Note that you must keep all the units the same as in G. The masses must be in kg and distance in meters.

The larger the planet and the larger the spaceship, the greater the pull of gravity. Also, the close the ship gets to the planet, the greater the gravity.

Note that since the Sun is huge, is causes a measurable pull on the mass of the ocean, even though the Sun is so far away.

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Weight inside and outside the body

Question

October 2, 2007

Does feces have the same weight inside the human body as it would outside the body, and how dies gravity affect it? jack - USA

14463

Answer

Assuming there is water and liquid in the intestines, feces would weigh slightly less in the body. Muscular contractions have more an effect on it than gravity.

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Where does gravity come from?

Question

December 24, 2006

Question: If you can measure the gravity effect, then can you tell me what gravity is or where the force comes from?

- USA

12685








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Answer

All matter possesses a force that attracts other matter to it. This force is called gravity. It is simply a property of matter, that is very important in holding the universe together.

G r r a v v i i t t y a c c c e l l e r r a t t i i o n o n E a r r t t h

We can show that the Universal Gravity Equation is the same as the standard equation for the force of gravity on Earth

F = mg where:

 m is the mass of the object



 g is the gravitational constant or acceleration due to gravity; g = 9.8 m/s² (meters per seconds squared) or 32 ft/s² mg is m times g

Special considerations

The acceleration due gravity on the Earth has been determined by experiments and measurement. It has also been shown that g is a constant.

But note that there are some special considerations in the equation F = mg . First of all, it is assumed that the object is relatively close to the Earth. This equation would not hold as well for objects higher than our weather satellites. Another assumption is that the Earth has a much greater mass than the object's mass m .

With these considerations in mind, we can show that F = GMm/R² is the same as F = mg when the object is relatively close to the Earth. The following material is the derivation of the simple gravity equation near the Earth from the Universal Gravity

Equation.

Mass

Let M equal the mass of the Earth. The approximate value for M = 6*10 24 kilograms (6 followed by 24 zeros). Also, let m be the mass of some object near the surface of the Earth. As you will see later, we don't need to know the specific mass of the object.

Distance is radius

This takes a little stretch of the imagination, but let's assume that an object near the surface of the Earth is attracted toward the center of the Earth, as if all of the Earth's matter was compressed at that point (as per Newton's assumption, when he came up with his theory). If r is the radius of the Earth plus a few meters, then the object near the surface would be a distance of r from the center of gravity.

The approximate radius of the Earth is 6.376*10 6 meters, so the distance between M and m is R = 6.376*10 6 m. Also, R² =

4*10 13 m² (meters -squared or square meters).








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Change units of G


Since a Newton is a kgm/s², we change the units o f G from Nm²/kg² to m 3 /s² -kg (meter-cubed per secondsquared-kilograms). This is done so that the units of G (m 3 /s² -kg) relates to the unit of m (kg) and units of g

(m/s²).

Calculation

Now let's put the values into the Universal Gravity Equation:

F = GMm /R² = (6.67*10 -11 m 3 /s²kg) * (6*10 24 kg) * m / (4*10 13 m²)

F = m*10 m/s²

Note how the various units will cancel out in the multiplication and division. This is important to verify that your units and the equation are correct.

Compare with g

Now we know that the force of gravity near the Earth is: F = mg

Thus, from the Universal Gravity Equation calculations above, g = 10 m/s². Since we used approximate values for r and M , that value is approximately g = 9.8 m/s² and the holds for the experimental values or measurements on Earth.







Document

Gravitation:

DESCRIPTIVE PART

1.

Newton’s’ Universal Gravitational Law:

Applications

2. Calculation of mass of the earth:

3. Calculation of density of the Earth:

4. Variation of “g” due to altitude:

5.

Variation of “g” due to depth:

6. Weightlessness in Satellites:

7. Artificial gravity:

Equations

Dimensions

PHYSICAL QUANTITY & SYMBOL DIMENSION UNIT

Short questions

Calculation of mass of sun:

How long for gravity to pull objects together?

What does the F stand for?

How does Einstein's theory apply to Earth?

Would a B-B and bowling ball fall at the same rate?

Gravity and a spaceship

Weight inside and outside the body

Where does gravity come from?

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Document

Gravitation:

DESCRIPTIVE PART

1.

Newton’s’ Universal Gravitational Law:

Applications

2. Calculation of mass of the earth:

3. Calculation of density of the Earth:

4. Variation of “g” due to altitude:

5.

Variation of “g” due to depth:

6. Weightlessness in Satellites:

7. Artificial gravity:

Equations

Dimensions

PHYSICAL QUANTITY & SYMBOL DIMENSION UNIT

Short questions

Calculation of mass of sun:

How long for gravity to pull objects together?

What does the F stand for?

How does Einstein's theory apply to Earth?

Would a B-B and bowling ball fall at the same rate?

Gravity and a spaceship

Weight inside and outside the body

Where does gravity come from?

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