Inertial and Gravitational Mass
Inertial Mass: – Inertial mass is defined as the mass of body by virtue of inertia of mass.
- By Newton’s Law F=ma
- m=F/a where m= inertial mass (as it is because of inertia of a body)
Gravitational Mass:-Gravitational mass is defined as the mass of the body by virtue of the gravitational force exerted by the earth.
- By Gravitation Force of attraction –
- F=GmM/r2
- m=Fr2/GM where
- F=GmM/r2
- m= mass of the object
- F=force of attraction exerted by the earth
- r=distance between object and earth
- M=mass of the earth
- Experimentally, Inertial mass= Gravitational mass
Problem: ( Class 11 Physics Gravitation )
Calculate the mass of the sun from the data given below:
Mean distance between Sun and Earth =1.5×1011m
Time taken by earth to complete one orbit around the sun = 1 year
Answer:
F= G MexMs/r2
Given: r= 1.5 x 1011m, v = 2π /r; T= 1 year =365 days =365x24x60x60sec
Fc= Mev2/r
F=Fc
GMeMs/r2 = Mev2/r = Me (2π r)2 /rT2
After calculation:-
Ms= 4n2r3/GT2
Ms=2×1030 kg.
Problem: ( Gravitation )
Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 105 ly.
Answer
Mass of our galaxy Milky Way, M = 2.5 × 1011 solar mass
Solar mass = Mass of Sun = 2.0 × 1036 kg
Mass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 ×1041 kg
Diameter of Milky Way, d = 105 ly
Radius of Milky Way, r = 5 × 104 ly
1ly = 9.46 × 1015 m
r =5 × 104 × 9.46 × 1015
= 4.73 ×1020 m
Since a star revolves around the galactic centre of the Milky Way, its time period is
given by the relation:
T= (4 π2r3/G M)1/2
= ((4 x(3.14)2x (4.73) 3x1060)/6.67×10-11x5x1041)1/2
= (39.48×105.82×1030/33.35)1/2
=(125.27×1030)1/2=1.12×1016 s
1year = 365x324x60x60s
1s=1/365x324x60x60 years
Therefore,
1.12×1016/365x324x60x60
=3.35×108 years.
Problem: – ( Gravitation )
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Answer
Orbital period of Io =Tlo = 1.769 days =1.769x24x60x60s
Orbital radius of Io = RIo = 4.22 × 108 m
Satellite Io is revolving around the Jupiter
Mass of the latter is given by the relation:
Mj=4 π2Rlo3/G Tlo2
Where,
Mj = Mass of Jupiter
G = Universal gravitational constant
Orbital radius of the Earth,
Te = 365.25 days = 365.25x24x60x60s
Orbital radius of the Earth,
Re = 1AU = 1.496×1011m
Mass of the Sun is given as:
Ms= 4 π2Re3/G Te2
Ms/ Mj = (4 π2Re3/G Te2) x (G Tlo2/4 π2Rlo3)
= (Re3/ Rlo3) x (Tlo2/ Te2)
= (1.769x24x60x60s/365.25x24x60x60s)2 x (1.496×1011m/4.22 × 108 m)
=1045.04
Ms/ Mj ~ 1000
Ms ~ 1000 Mj
Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.
Problem: – ( Gravitation )
How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km.
Answer:
Orbital radius of the Earth around the Sun, r = 1.5 × 1011 m
Time taken by the Earth to complete one revolution around the Sun,
T = 1 year = 365.25 days= 365.25 × 24 × 60 × 60 s
Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2
Thus, mass of the Sun can be calculated using the relation
M=4 π2r3/GT2
= (4x (3.14)2x(1.5×1011)3)/(6.67×10-11 x(365.25 × 24 × 60 × 60))
=133.24×10/6.64×104
=2.0×1030Kg
Hence, the mass of the Sun is 2 × 1030 kg.