# Chapter 8 : Gravitation¶

## Example 8.1 , page : 185¶

In [1]:
# Importing module

import math

# Variable declaration

mp=1                     # For convenience,mass is assumed to be unity
rp=1                     # For convenience,sun-planet distance at perihelton is assumed to be unity
vp=1                     # For convenience,speed of the planet at perihelton is assumed to be unity
ra=1                     # For convenience,sun-planet distance at aphelton is assumed to be unity
va=1                     # For convenience,speed of the planet at aphelton is assumed to be unity
Lp=mp*rp*vp              # Angular momentum at perihelton
La=mp*ra*va              # Angular momentum at ahelton

# Result

# From angular momentum conservation, mp*rp*vp = mp*ra*va or vp/va = rp/ra
# From Kepler’s second law, equal areas are swept in equal times
print(" The planet will take a longer time to traverse BAC than CPB")

 The planet will take a longer time to traverse BAC than CPB


## Example 8.2 , page : 187¶

In [2]:
# Importing module

import math

# Variable declaration

G=6.67*pow(10,-11)       # Gravitational constant
m=1                      # For convenience,mass is assumed to be unity
x=30                     # The angle between GC and the positive x-axis is 30° and so is the angle between GB and the negative x-axis
a=math.cos(y)
b=math.sin(y)
v1=(0,1,0)
v2=(-a,-b,0)
v3=(a,-b,0)
c=(2*G*pow(m,2))/1       # 2Gm²/1

# Calculation

#(a)
F1=[y * c for y in v1]   # F(GA)
F2=[y * c for y in v2]   # F(GB)
F3=[y * c for y in v3]   # F(GC)
# From the principle of superposition and the law of vector addition, the resultant gravitational force FR on (2m) is given by
Fa=[sum(x) for x in zip(F1,F2,F3)]

#(b)
# By symmetry the x-component of the force cancels out and the y-component survives
Fb=4-2                   # 4Gm² j - 2Gm² j

# Result

print("(a) The force acting =",Fa,"≈ 0")
print("(b) The force acting =",Fb,"Gm²")

(a) The force acting = [0.0, 2.5849394142282115e-26, 0.0] ≈ 0
(b) The force acting = 2 Gm²


## Example 8.3 , page : 192¶

In [3]:
# Importing module

import math

# Variable declaration

G=6.67*pow(10,-11)       # Gravitational constant
m=1                      # For convenience,mass is assumed to be unity
l=1                      # For convenience,side of the square is assumed to be unity
c=(G*pow(m,2))/l
n=4                      # Number of particles

# Calculation

d=math.sqrt(2)
# If the side of a square is l then the diagonal distance is  √2l
# We have four mass pairs at distance l and two diagonal pairs at distance √2l
# Since the Potential Energy of a system of four particles is -4Gm²/l) - 2Gm²/dl
w=(-n-(2/d))
# If the side of a square is l then the diagonal distance from the centre to corner is
# Since the Gravitational Potential at the centre of the square
u=-n*(2/d)

# Result

print ("Potential energy of a system of four particles =",w,"Gm²/l")
print("The gravitational potential at the centre of the square =",u,"Gm²/l")

Potential energy of a system of four particles = -5.414213562373095 Gm²/l
The gravitational potential at the centre of the square = -5.65685424949238 Gm²/l


## Example 8.4 , page : 193¶

In [4]:
# Importing module

import math

# Variable declaration

R=1                      # For convenience,radii of both the spheres is assumed to be unity
M=1                      # For convenience,mass is assumed to be unity
m1=M                     # Mass of the first sphere
m2=6*M                   # Mass of the second sphere
m=1                      # Since the mass of the projectile is unknown,take it as unity
d=6*R                    # Distance between the centres of both the spheres
r=1                      # The distance from the centre of first sphere to the neutral point N

G=6.67*pow(10,-11)       # Gravitational constant

# Calculation

# Since N is the neutral point; GMm/r² = 4GMm/(6R-r)²  and we get
r=2*R
# The mechanical energy at the surface of M is; Et = m(v^2)/2 - GMm/R - 4GMm/5R
# The mechanical energy at N is; En = -GMm/2R - 4GMm/4R
# From the principle of conservation of mechanical energy; Et = En and we get
v_sqr=2*((4/5)-(1/2))

# Result

print("Minimum speed of the projectile to reach the surface of the second sphere =","(",round(v_sqr,5),"GM/R",")","^(1/2)")

Minimum speed of the projectile to reach the surface of the second sphere = ( 0.6 GM/R ) ^(1/2)


## Example 8.5 , page : 195¶

In [5]:
# Importing module

import math

# Variable declaration

π=3.14                   # Constant pi
G=6.67*pow(10,-11)       # Gravitational constant
R=9.4*pow(10,3)          # Orbital radius of Mars in km
T=459*60
Te=365                   # Period of revolution of Earth
r=1.52                   # Ratio of  Rms/Res, where Rms is the mars-sun distance and Res is the earth-sun distance.

# Calculation

# (i)
R=R*pow(10,3)
# Using Kepler's 3rd law:T²=4π²(R^3)/GMm
Mm=(4*pow(π,2)*pow(R,3))/(G*pow(T,2))

# (ii)
# Using Kepler's 3rd law: Tm²/Te² = (Rms^3/Res^3)
Tm=pow(r,(3/2))*365

# Result

print("(i)  Mass of Mars =",Mm,"kg")
print("(ii) Period of revolution of Mars =",Tm,"days")

(i)  Mass of Mars = 6.475139697520706e+23 kg
(ii) Period of revolution of Mars = 684.0033777694376 days


## Example 8.6 , page : 195¶

In [6]:
# Importing module

import math

# Variable declaration

g=9.81                   # Acceleration due to gravity
G=6.67*pow(10,-11)       # Gravitational constant
Re=6.37*pow(10,6)        # Radius of Earth in m
R=3.84*pow(10,8)         # Distance of Moon from Earth in m
T=27.3                   # Period of revolution of Moon in days
π=3.14                   # Constant pi

# Calculation

# I Method
# Using Newton's 2nd law of motion:g = F/m = GMe/Re²
Me1=(g*pow(Re,2))/G

# II Method
# Using Kepler's 3rd law: T²= 4π²(R^3)/GMe
T1=T*24*60*60
Me2=(4*pow(π,2)*pow(R,3))/(G*pow(T1,2))

#Result

print("Mass of the Earth =",Me1,"kg")
print("Mass of the Earth =",Me2,"kg")

Mass of the Earth = 5.967906881559221e+24 kg
Mass of the Earth = 6.017752855396305e+24 kg


## Example 8.7 , page : 195¶

In [7]:
# Importing module

import math

# Variable declaration

k=pow(10,-13)           # A constant = 4π² / GME
Re=3.84*pow(10,5)       # Distance of the Moon from the Earth in m

# Calculation

k=pow(10,-13)*(pow(1/(24*60*60),2))*(1/pow((1/1000),3))
T2=k*pow(Re,3)
T=math.sqrt(T2)         # Period of revolution of Moon in days

# Result

print("Period of revolution of Moon =",round(T,1),"days")

Period of revolution of Moon = 27.5 days


## Example 8.8 , page : 196¶

In [8]:
# Importing module

import math

# Variable declaration

m=400                   # Mass of satellite in kg
Re=6.37*pow(10,6)       # Radius of Earth in m
g=9.81                  # Acceleration due to gravity

# Calculation

# Change in energy is E=Ef-Ei
ΔE=(g*m*Re)/8           # Change in Total energy
# Since Potential Energy is twice as the change in Total Energy (V = Vf - Vi)
ΔV=2*ΔE                 # Change in Potential Energy in J

# Result

print("Change in Kinetic Energy =",round(ΔE,4),"J")
print("Change in Potential Energy =",round(ΔV,4),"J")

Change in Kinetic Energy = 3124485000.0 J
Change in Potential Energy = 6248970000.0 J