## Saturday, 22 December 2012

$$P_{saturn} = 4\pi R_{Saturn}^{2}\sigma T_{Saturn}^{4}$$

First calculate with T as 76K:

$$P_{Saturn}^{T=76K} = 4\pi×57500^2×5.6703*10^{−8}×(76)^4$$
$$= 7.8596*10^{10} W$$

Now T as 93K:

$$P_{Saturn}^{T=93K} = 4\pi×57500^2×5.6703*10^{−8}×(93)^4$$
$$= 1.7623*10^{11}W$$

We have our two powers one $P_{Saturn}^{T=93K}$ and one $P_{Saturn}^{T=76K}$ so now we find the difference:

$$P_{difference} = P_{Saturn}^{T=76K} - P_{Saturn}^{T=93K} = 9.7634*10^{10}W$$

So Saturn radiates 0.554 times the power that saturn absorbes by sun which increases the temp to 93K.

## Thursday, 20 December 2012

### Sun and Saturn

Some info on Sun And Saturn:

$E_{\odot} = 4\pi{}R_{\odot}^2\sigma{}T_{\odot}^4$

$E_{\odot} = 4*\pi{}*(695.508*10^6)^2*5.9703*10^-8*5777^4$
$= 3.88391*10^{26} W$

At Saturn, this energy is passing through a sphere with a radius of $R_{sphere}$, the distance between Saturn and the Sun, and the energy passing through each square metre of the sphere is given by
$Power density_\text{Rsphere} = \frac{E_{\odot}}{4\pi*R_\text{sphere}^2}$

$\text{Power density}_\text{Rsphere} = \frac{3.88391*10^{26}}{4*\pi*(9.6AU * 149.6*10^9m/AU)}$
$\text{Power density}_\text{Rsphere} = 2.152070365*10^{13}W/m^2$

$E_{absorbed} = \pi R_{saturn}^2*Energy_{Rsphere}$

$E_{absorbed} = \pi*(58232)^2*2.15207*10^{13}$
$E_{absorbed} = 2.2926*10^23km^2W$

## Saturday, 15 December 2012

### Roche limit Questions

Find the roche limit of a planet with Mass $5.87*10^28Kg$ (Heehee! I love to exaggerate some things!) and me with a mass of $45Kg$. My radius is about 13cm or 0.00013km! The Formula is this for a rigid object $d = 2.44\; R_m\left( \frac {M_M} {M_m} \right)^{\frac{1}{3}}$

Ans:

$M_M = 5.87*10^28Kg$

$M_m = 45Kg$

$R_m = 0.00013km$

$d = 2.44*R_m(\frac{M_M}{M_m})^\frac{1}{3}$

$d = 2.44*0.00013*(\frac{5.87*10^28}{45})^(\frac{1}{3})$

$d = 346583.8921\text{km or} 3.465*10^5 km$ #at this distance I will be the planets ring!

New question:

A supermassive star with density $7005.5Kg/m^2$ and a planets density of $1000Kg/m^2$ and $R = 47008Km$. The planet is liquidy, find the roche limit with the formula $d \approx 2.44R\left( \frac {\rho_M} {\rho_m} \right)^{1/3}$

Ans:

$p_M = 7005.5kg.m^2$

$p_m = 1000kg/m^2$

$R = 47008Km$

$d = 2.44 * R(\frac{p_M}{p_m})^(\frac{1}{3})$

$d = 2.44 * 47008(\frac{7005.5}{1000})^(frac{1}{3})$

$d = 219469.7Km = 2.19*10^5km$

### Roche limit

Roche limit is a limit at which a planet tears apart a moon or object via tidal Forces

The formula for Roche Limit for a rigid object is $d = 2.44\; R_m\left( \frac {M_M} {M_m} \right)^{\frac{1}{3}}$

Where $M_M$ is The mass of the primary or the planet.
$M_m$ is the mass of the secondary or moon.
$R_m$ is the Radius of the secondry or moon.

The formula for fluid objects is $d \approx 2.44R\left( \frac {\rho_M} {\rho_m} \right)^{1/3}$
Where $R$ is the Radius of the primary or planet
$P_M$ is the Density of the primary or planet.
$P_m$ is the density of the secondry or moon. Consider an orbiting mass of fluid held together by gravity, here viewed from above the orbital plane. Far from the Roche limit the mass is practically spherical. Closer to the Roche limit the body is deformed by tidal forces Within the Roche limit the mass's own gravity can no longer withstand the tidal forces, and the body disintegrates. Particles closer to the primary move more quickly than particles farther away, as represented by the red arrows. The varying orbital speed of the material eventually causes it to form a ring.

## Friday, 14 December 2012

### Which is spookiest

For the new poll Which is Spookiest refer here here

## How long would it take to travel from one end of the visible Universe to the other end, if you were travelling at the different speeds ???

Ans:

$$1. speed = 1 Ly/sec$$

$$\text{Distance from one edge to the other}= 93*10^9Ly$$

$$Time = \frac{distance}{speed}$$

$$Time = \frac{93*10^9}{1} = 93*10^9seconds = 1.55*10^9min = 25.83*10^6hours$$

$$=1.076*10^6days = 2948.63years$$

Now:

$$2.speed = 1parcsec/sec$$

$$\text{distance from one edge to other}= 28*10^9parcsec$$

$$Time = \frac{28*10^9}{1} = 28*10^9sec = 4.66*10^8min = 7.77*10^6hours = 3.2*10^5days$$

$$= 876.712years$$

## Thursday, 13 December 2012

### Python codes for easy work

>>> import math
>>> def grav_poten_energy(m1,m2,r):
constant = 6.672e-11
form = -((constant*m1*m2)/r)
return form

>>> def kinetic_energy(m, v):
form = (1/2)*m*v**2
return form

>>> def angular_momentum(m, v, R):
return m*v*R

>>> def momentum(m, v):
return m*v

>>> def centripetal_acceleration(v, R):
return (v**2/R)

>>> def gravity(G, m, r):
g = 6.672e-11
if G != g:
return None
form = (G*m)/r**2
return form

>>> def gravity_two_object(G, m1, m2, r):
g = 6.672e-11
if G != g:
return None
form = (G*m1*m2)/r**2
return form

>>> def Kepler_third_law2(K, R):
form=(4*math.pi**2)/(K*R)
return form

>>> def Kepler_third_law(K, a):
form = math.sqrt(K*a**3)
return form

>>> def synodic_period(period1, period2):
form1 = (period1*period2)/(period2-period2)
return form1

>>> def resolution(arcsec, distance):
degree = arcsec/3600
form2 = (math.pi/180) * distance
return degree*form2

>>> def alt(zen):
form = 90 - zen
return form

>>> def zenith(lat, dec):
return lat - dec

>>> def distance(a,r):
form1 = a/57.3
main = form1 * r
return main

>>> def flux(sigma, T):
return sigma*T**4

>>> def luminosity(R, F):
return 4*math.pi*R**2*F

>>> def doppler_effect(v, c, lambda0):
return lambda0*(1-(v/c))

## Sun massed black-hole?

* Q.1. Calculate the radius of a black hole that has the same Mass of Sun.

Ans: $r_\mathrm{s} = \frac{2Gm}{c^2},$     this is our Formula
$$m =\text{ Mass of the sun} = 1.989x10^30 kg$$

$$G =\text{ gravitational constant} = 6.672x10^-11 Nm^2/kg^2$$

$$c = 2.9979x10^8 m/s$$
so:

$$R =2 * 6.672x10^-11 * 1.989x10^30/(2.9979x10^8)^2$$

$$R = ( 2.65412x10^20 ) / ( 8.98740441x10^16 )$$

$$R = 2953.156973 m or 2.953 km$$

* I am currently reading this book  "voyages through the universe ,third edition by Frankoi ,Morrison ,Wolff ",this book has some exercise which have no answers ,here I am solving the question on my own for knowledge .

## Thursday, 29 November 2012

### resolution

The resolution of a telescope, as we will learn, is the minimal angular size of features it can resolve. If we have a telescope with a resolution of 2'' (2 arc-seconds) what is the size, in kilometers (km), of the smallest crater we can make out on the moon, at a distance of 384,400 km? (The Moon's diameter is 3480 km, for comparison.) Round your answer to two significant figures.

ANS:

a = 2"
R = 384400km

a/(360*3600)*(2pi*R)
2/(360*3600)*(2pi*384400)    #I multiply 360 with 3600 for coversion to arcsec
= 3.7272km

there is another way:

(a/3600)*(pi/180)*R
(2/3600)*(pi/180)*384400
=3.7272km

## Wednesday, 21 November 2012

### A thought?

(with a serious face and the background music is as if you have found an invention, you wonder)

"Sometimes I wonder?" (music stops) "HOW THE HECK AM I SITTING OVER THERE WHEN I AM STANDING OVER HERE!!!!!!!!!!!!!!!!!!!!!"

## Saturday, 10 November 2012

### event horizon

I am going to introduce you guys to a new formula or concept, called the schwarzschild radius it is to find the event horizon of a black hole. It depends on the mass of the black hole. bigger mass bigger radius.

Formula is:

R = (2xGxM)/c^2
do not beleive in this formula while looking, think why, how and then believe in it.

related questions on there way.

## Q.7 what is the escape velocity of the neutron star in question 6?

Ans:
m = 3.96x10^30 Kg
r = 10Km
G = 6.67x10^-11 N/kg^2

formula = (2 x G x m)/r^2
= (2 x 6.67x10^-11 x 3.96x10^30) / 10^2
= 5.28x10^18 km/s

### Question 6

Q.6 Take a neutron star with 2Msun but a radius of 10Km! What is the accelaration of gravity at the surface of it? how much greater than g(9.8) is it? Finally how much will you weigh on it?(assume you somehow dont turn into a puddle of protoplasm)?

Ans:

sun mass = 1.98x10^30
m = 2 x 1.98x10^30 = 3.96x10^30 Kg
G = 6.67x10^-11 N/Kg^2
m2 = my weight = 40Kg (assuming)
g = 9.8 m/s^2

formula = (G x m)/r^2
= (6.67x10^-11 x 3.96x10^30) / 10^2
=  2.64132xx10^18 km/s^2
so:

Gravity = 2.64x10^18 km/s^2
g = 9.8 m/s^2

times greater = gravity/g
= 2.64x10^18 / 9.8
= 2.69x10^17 times   g
I would weigh:

m2 = 40 Kg
F = m x gravity
= 40 x 2.64x10^18
= 1.055x10^20 Kg !!!!!!!!!!!!!!

## Sunday, 21 October 2012

Look at this 360 panorama of Mars by Curiosity:

http://www.360cities.net/image/curiosity-mars

## Saturday, 20 October 2012

### binry destruction

This is inter_activity which should be done in a blog. I will ask a problem, we will discuss, and then solve it.

This one interested me so here it is:

A. The star Sirius (the brightest star in the sky) has a white dwarf companion. Sirius has a mass of about
2  Msun and it is still on the main sequence, while its companion is already a star corpse. Remember that
a white dwarf cant have a mass greater  than 1.4 Msun. (Chandrasekhar limit http://en.wikipedia.org/wiki/Subrahmanyan_Chandrasekhar ). Assuming that they both formed on the same
time have a group discuss how Sirius could have a white dwarf companion. HINT: Was the initial mass of                                               the white dwarf star more or lesser than Sirius?

### The World From The Eyes of Muhammad Shaheer Niazi: Hi ,I am Muhammad Shaheer Niazi ,welcome to my blo...

The World From The Eyes of Muhammad Shaheer Niazi: Hi ,I am Muhammad Shaheer Niazi ,welcome to my blo...: Hi ,I am Muhammad Shaheer Niazi ,welcome to my blog. I am 12 years old ,my interests are astronomy ,Maths ,physics ,history ,biology and ch...

### NASA new finds on galaxies formation

 Simulations such as this will help astronomers better understand the new findings in galaxy evolution. It tracks the development of a single disk galaxy from shortly after the Big Bang to the present day. Colors reveal old stars (red), young stars (white and bright blue) and the distribution of gas density (pale blue);

credit NASA
http://forums.udacity.com/ph100/questions/9173/
my-own-gravity-at-10cm-away-from-me-hehehe-and-
now-black-holes-gravity-to-test-if-light-can-e
scape-one-or-not-and-again-some-more-11-questions-
for-help-in-the-black-hole-question

**Q4. Say a white dwarf has a mass of the sun but the radius of earth.
What is the surface gravity of it? How much will you weigh on it??**

so:

mass = 1.98x10^30 kg
gravitational constant = 6.67x10^-11 Nm/kg^2
a = (G x m)/r^2
= (6.67x10^-11 x 1.98x10^30) / 6.4x10^6^2
= 3224267.5 m/s^2
we will wheigh:

m=40kg
a = 3.22x10^6 m/s^2

f=ma
=40 x 3.22x10^6
=1.28x10^8 N !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

## Friday, 19 October 2012

### The python version

Q.1

import math

G = 6.67e-11
if constant != G:
return "INCORRECT, >:)"
formula1 = (constant * mass)
main = formula1 / formula2
if constant == G:
return "YOU GOT IT RIGHT ,correct", main
Q.2

def race_with_light_or_a_fly(acceleration,s_light):
light = 300000000 #m/s
if s_light>acceleration:
return "Zap, Ill never let you go you fake programmer!", "P.S its
light and no"
one _meter = 1.057e-16 #LY
if s_light<acceleration:
return "yoyoyoyoyoyo!!!!, light loves you, you and light are going in blackhole"

print cosmic_gravities(40,0.1,6.67e-11)

print cosmic_gravities(1.99e3,5,6.67e-11)
print cosmic_gravities(11212,787,9e-11)

light = 300000000

print race_with_light_or_a_fly(5.7e6,light)
print race_with_light_or_a_fly(5.7e9,light)
Q.3

con = 6.67e-11 #Nm/kg
if constant == con:
return "I wanna move it move it , correct", math.sqrt(formulae)/1000
if constant != con:
return "Im falling, im going down, incorrect", False

G = 6.67e-11
print escape_velocity(6.67e-11,79997977979,0.009)
print escape_velocity(6.67e-11,5.34e24,6.4e6)
print escape_velocity(7e8,5555,55555)
Collapse this post

### The World From The Eyes of Muhammad Shaheer Niazi: My solutions to the Questions written in the book ...

The World From The Eyes of Muhammad Shaheer Niazi: My solutions to the Questions written in the book ...: This I have done for more help in the black hole question:: How much would you weigh on a stellar corpse? Lets calculate what is it rea...

### My solutions to the Questions written in the book .

This I have done for more help in the black hole question::
How much would you weigh on a stellar corpse? Lets calculate what is it really lie near a white dwarf or Neutron star. First, lets ask how strongly gravity accelerates you on a body (which is a measure of how strongly you are pulled to the surface). the force between two bodies is calculated as
     F=(G x M1 x M2)/r^2

Where G is the gravitational constant --- 6.67x10^-11 Nm/kg^2, M1 M2 are the masses of two bodies, and r is there seperation. Also, froom Newtons law,
     F = M1 x a

Where a is the acceleration of a body with mass M. So lets consider a situation of a mass - say, you- standing on a body, such as the Earth or a white dwarf. You are M1 and the body you are standing on is M2. The distance between the center of gravity of the body on which you stand is r. The force exerted on you is
     F = M1 x a = (G x M1 x M2) / r^2

The acceleration of gravity on that world which is given as the letter g, is (G x M)/r^2. We can now calculate the acceleration of Gravity for our Earth:
     g=(6.67x10^-11 x 6x10^24) / (6.4x10^6)

=9.8 m/s^2

The speed you need to get away from a body is escape velocity. It is given by the formula:
     v^2=(2 x G x M) / r

For the Earth, v^2 = 2*(6.67x10^-11 x 6x10^24) / 6.4x10^6 = 1.25 x 10^8. Now remembering this is the square of the escape velocity. Taking the square root gives us 11.2 km/s. This is the speed(25,000 mph) that you have to go to escape Earths gravity. The density of a body is its mass over volume. The volume of a sphere is V = 4/3 x pi x r^3, so
     d = M/((4/3) x pi x r^3)
= 6x10^24 / (1.33 x 3.14 x (6.4x110^6))
= 5500 KJg/m^3

This explanation I took from My great book voyages through the universe 3rd edition from the chp22 Questions.http://www.textbooks.com/ISBN/9780534409050/Andrew-Fraknoi-David-Morrison-and-Sidney-Wolfe/Voyages-Through-the-Universe-and-With-CD-ROM_-_0534409059.php
Now ill answer the questions (figuring out yourself) myself. Aswers are not given in the book.
Q1. What is the acceleration of gravity at the surface of the sun!?How much greater is it than Earths? How much would you weigh on the sun?
My Ans:
              Mass of sun = 1.98x10^30 kg
r = 6.96x10^8 m
G = 6.67x10^-11 Nm/kg

so:
              g = (G x M)/r^2
= (6.67x10^-11 x 1.98x10^30)/6.96x10^8^2
= 272.6 m/s^2

How much greater:
              Earth= 9.8 m/s^2
sun= 272.8 m/s^2

greater = 272.8 / 9.8
= 27.8 times the earth

How much will I weigh:
              me = 40 kg
sun g = 272.8
wheight = 272.8 x 40 = 10912N                  #OMG! I weigh 1.1 tonne

Q2. What is the escape velocity of the sun? How much is it greater than Earths?
My Ans:
               G = 6.67x10^-11 Nm/kg
M = 1.98x10^30 kg
r = 6.96x10^8 m

so:
               v = sqrt(2 x G x M)/r
= 616035.7 m/s = 616 km/s              # nice

Q3.what is the density of the sun? My Ans:
              V = 4/3 x pi x r^3
d = M/(4/3) x pi x r^3

v = 4/3 x 3.142 x 6.96x10^8
= 1.412x10^27m = 1.412x10^24km

d = 1.98x10^30 / 1.412x10^24
= 1.3943kg/m^2
Hi ,I am Muhammad Shaheer Niazi ,welcome to my blog. I am 12 years old ,my interests are astronomy ,Maths ,physics ,history ,biology and chemistry .