User:Ravi Mistry r^2

Hello

"I inspire and motivate by one of the great scientist albert einstein during my infancy from books and tv shows.He is my favorite scientist.I am so much interested in theoretical physics.I refer many books relative that like special and general theory of relativity,a brief history of time,quantum mechanics and much more.Before i died i want to take a journey of time travel in past.it my dream.I will really take that journey in reality its my pledge to my self."

                                                      May be secret

according to our scientist we acn run fast more than the speed of light  but there is requirement of more and more energy ..............we can run fast more than the speed of light through the warp drive ...

according to newton

F=Gm1m2/re^2

so we can write

F=Gm1m2/^2

if r=0

so F=Gm1m2/(0)^2

     =infinite

so F=infinite

so F=ma

therefore a=infinite

and according to albert einstein

E=mc^2

so we can write for any particle that

E=mv^2

then differentiation realtive to time

therefore

                          dE/dt=dmv^2/dt

therefore dE/dt=m dv^2/dt

therefore dE/dt=m2v dv/dt

therefore dE/dt=2mva

so dE/dt=2mv(infinite) (because there is a=infinte)

therefore dE/dt=infinite

we can use this energy for warp drive we can compress our froward space and behind space decompress that the think of warp drive but there requirement infinite energy

may be i am right

                                                           Physics And Maths

Whats the solution of     r limitt to 0 Gm1m2/r^2 (i want to know this )

also i want to know n limit to 0 0/n=?

and n limit to 0 infinite/n=?

new topic

according to einstein theory of relativity

E=mc^2 so we can write for any particle that

E=mv^2

then we can differentiation of this formula relative to time

so that si

                            dE/dt=dmv^2/dt

threfore dE/dt=mdv^2/dt (because here we beleive that mass contant)

therefore dE/dt=m2vdv/dt

therfore dE/dt=m2va (because there is we can write a (acceleration ) instead of dv/vt )

threfore dE/dt=2mva

therefore dE/dt=2Fv (because there is F=ma)

also we can write dE/dt=2mva

threfore dE/dt=2pa

mass (m) change with velocity and velocity chnge with time

M=M0/sqrt 1-v^2/c^2 (according to theory of relartivity )

if velocity chand with also time so mass also change with time

so M=M0/sqrt 1-v^2/c^2

       =M0*c/sqrt c^2-v^2

then differentiation relative to velocity v

so dm/dv=M0*c d1/sqrt c^2-v^2

                =M0*c  d(c^2-v^2)^-1/2

                =M0*c [(-1/2(c^2-v^2)^-3/2)  d(c^2-v^2)/dt]

                =M0*c [-1/2(c^2-v^2)^-3/2     (-2v)   dv/dt]

                =M0*c [(c^2-v^2)-3/2  va]     (because there is dv/dt=a)

               =M0*c *v*a (c^2-v^2)

now,

dm/dt*dt/dv=dm/dv

so dm/dt*dt/dv=M0cva(c^2-v^2)^-3/2

therefore

dm/dt=M0cva(c^2-v^2)^-3/2 dv/dt

          =M0cva^2(c^2-V^2)^-3/2

         =M0cva^2/(c^2-v^2)^3/2

                               A REAL FACT ABOUT OUR EARTH AND ANYTHING ELSE SUCH AS PLANETS OR STARS..

WHEN ANYTHING TRAVEL AT TO SPEED OF LIGHT ITS MASS INCREASE ACCORDING TO EINTSEIN SPECIAL THEORY OF RELATIVITY

SUPPOSE OUR EARTH GET TO AT CLOSE TO SPEED OF LIGHT SO ITS MASS INCREASE BUT HOW MUCH?

NOW CALCULATE..

ACCORDING TO EINSTEIN SPECIAL THEORY OF RELATIVITY,

M=M0/sqrt 1-v^2/c^2

WHERE M=MASS ACCORDING TO SPEED

M0=INITIAL MASS

v=SPEED OF ANY THING OR ANYBODY

c=SPEED OF LIGHT

HERE.....M0=6*10^24 (MASS OF THE EARTH)

v=1,06,817 km/h (SPEED OF OUR EARTH ROUND AND ROUND TO THE SUN)(NOW SPEED)

CALCULATION OF VELOCITY ,

OUR EARTH DISTANCE FROM THE SUN d=1.49*10^8 km

(HERE d=r=1.49*10^8 km)

SO ITS ROUND PATH DISTANCE, d=2*3.14*r

=2*3.14*1.49*10^8 km =9.35*10^8 km

AND TIME FOR COVER DISTANCE t=365 DAYS=8760 HOURS SO v=d/t

=9.35*10^8/8760

=1,06,817 km/h

=29.617 km/h

SUPPOSE OUR EARTH GET TO CLOSE TO SPEED OF LIGHT LIKE 2,99,791.999999 km/h (BECAUSE DARK ENERGY)

SO v=2,99,791.999999 km/h

AND c=2,99792 km/h

SO M=6*(10^24)/sqrt 1-(299791.999999/299792)^2

=6*(10^24)/sqrt 1-(0.99999999999)^2

=6*(10^24)/sqrt 1-0.99999999998

=6*(10^24)/sqrt 0.00000000002 =6*(10^24)/0.00000447213

=13.14*10^29 kg

1) FIRST SITUATION

IN THIS CASE SUPPOSE OUR EARTH RADIUS CONSTANT SO,

ACOORDING TO ANGULAR MOMENTUM,

L1=L2

I1W1=I2W2 (W IS OMEGA)

2/5M1R^2W1=2/5M2R^2W2 (BECAUSE HIGH DENSITY ROUND SHAPE MOMENTUM OF INERTIA IS 2/5MR^2)

M1W1=M2W2

M1*2*3.14*f1=M2*2*3.14*f2 (BECAUSE W=2*3.14*f)

M1/T1=M2/T2 (BECAUSE f=1/T)

HERE M1=NOW MASS

T1=NOW PERIODIC TIME M2=ON THE SECOND POSITION MASS

T2=ON THE SECOND POSOTION PERIODIC TIME

SO FOR THE EARTH,

M1=6*10^24

T1=24 HOURS (ITS ON OWN AXIS ROTATION TIME)

M2=13.41*10^29 kg

T2=HOW MUCH ?

NOW CALCULATE,

6*10^24/24=13.41*(10^29)/T2

THEREFORE T2=13.41*(10^29)*24/6*10^24

=13.41*4*(10^5)/1

=53.64*10^5 HOURS

=2,23,500 DAYS

=612.32 YEAR ITS ON OWN AXIS ROTATION TIME.

(THIS IS ONLY FOR ROUND MOTION)

2)SECOND SITUATION

IN THIS CASE SUPPOSE OUR EARTH RADIUS 1/4 DECREASE LESS THAN NOW RADIUS.

AND ALSO MASS AS FIRST SITUATION,

SO ACCORDING TO ANGULAR MOMENTUM,

L1=L2

I1W1=I2W2 (W IS OMEGA)

2/5M1R1^2W1=2/5M2R2^2W2 (BECAUSE HIGH DENSITY ROUND SHAPE MOMENT OF INERTIA IS 2/5 (BECAUSE HIGH DENSITY ROUND SHAPE MOMENT OF INERTIA IS 2/5M R^2)

M1R1^2W1=M2R2^2W2

M1R1^2*3.14*f1=M2R2^2*3.14*f2 (BECAUSE W=2*3.14*f)

M1R1^2/T1=M2R2^2/T2 (BECAUSE f=1/T)

HERE, M1=NOW MASS

T1=NOW PERIODIC TIME

R1=NOW RADIUS

M2=ON THE SECOND POSOTION MASS

T2=ON THE SECOND POSOTION PERIODIC TIME

R2=ON THE SECOND POSITION RADIUS

SO FOR THE EARTH,

M1=6*10^24

T1=24 HOURS (ITS ON OWN AXIS ROTATION TIME)

R1=6400 km

M2=13.41*10^29 kg

R2=6400/4 km

T2=HOW MUCH ?

NOW CALCULATE,

6*10^24*(6400)^2/24=13.41*(10^29)*(6400/4)^2/T2

THEREFORE T2=13.41*(10^29)*(6400/4)^2*24/6*10^24*(6400)^2

=13.41*(10^29)*24/6*(10^24)*16

=13.41*(10^5)*4/16

=13.41*(10^5)/4

=3.35*10^5

=335000 HOURS

=13985.333 DAYS

=38.242008 YEARS

NOW REAL FACT IS THAT EARTH MOVE FOUR TYPE MOTION...

1)ON ITS AXIS

2)ROUND TO THE SUN

3)ROUND TO THE CENTER OF GALAXY

4)LINEAR MOTION

RADIUS OF EARTH r=6400 km SO d=2*3.14*r=2*3.14*6400=40192 km ON ITS AXIS MOTION DISTANCE COMPLETE TIME t=24 hr

SO v=d/t=40192/24=1674.66 km/h=0.46 km/s(we can use its own axis speed because precession)............................................................................(1)

ROUND TO THE SUN SPEED =1,06,817 km/h

CALCULATION ,

OUR EARTH DISTANCE FROM THE SUN d=1.49*10^8 km

(HERE d=r=1.49*10 km)

SO ITS ROUND PATH DISTANCE,

d=2*3.14*r

=2*3.14*1.49*10^8 km

=9.35*10^8 km

AND TIME FOR COVER DISTANCE

t=365 DAYS

=8760 HOURS

SO v=d/t

=9.35*10^8/8760

=1,06,817 km/h

=29.671 km/s ...................................................................................................................(2)

OUR EARTH AND SUN DISTANCE FROM THE CENTER OF GALAXY

d=30,000 LIGHT YEAR

(HERE d=r=30,000 LIGHT YEAR)

SO ITS ROUND PATH DISTANCE ,

d=2*3.14*r

=2*3.14*30000*9.46*10^12 km (BECAUSE 1 LIGHT YEAR =9.46*10^12 km)

=178.2264*10^16 km

AND TIME FOR COMPLETE DISTANCE,

t=22*10^7 YEAR

=8030*10^7 DAYS

=192720*10^7 HOURS

SO v=d/t

=178.2264*10^16/192720*10^7

=178.2264*10^16/19.2720*10^11

=178.2264*10^5/19.2720

=9.247*10^5 km/h

=0.0025686*10^5 km/s

=256.86 km/s .......................................................................................................................(3)

AND OUR EARTH LINEAR SPEED IS SUPPOSE 5000..........................................................................(4) ITS ACCORDING TO SIMPLE LOGIC.

ONE GALAXY WHICH NAME IS 3C273 ITS SPEED MASURE BY TELESCOPE IN 1962 IS 47,400 km/s WHICH AWAY FROM US 2*10^9 LIGHT YEAR , AND SECOND GALAXY WHICH NAME IS BR 1202-07 ITS SPEED MASURE BY TELESCOPE IN 1991 IS 2,70,000 km/s WHICH AWAY FROM US 10*10^9 LIGHT YEAR.

SO NOW OUR EARTH OUR EARTH TOTAL EFFECTIVE SPEED IN OUR THIS CALCULATION IS

TOTAL SPEED =5000+256.86+29.671+0.46

=5286.991 km/s

(IT IS APPROXIMATE ,IT IS NOT REAL SPEED)

NEXT FOLLOW FIRST AND SECOND SITUATION............