A gravitational field is a region of space surrounding a body that has the property of mass. Sir Isaac Newton, in 1666, propounded the universal law of gravitation.
The law states that, ” the force of attraction between two given particles of masses M and m is directly proportional to the product of the masses and inversely proportional to the square of their distance of separation”.
Where Fg is the gravitational force in Newton,N,
Gis the universal or gravitational constant of value 6.7 × 10-11 and expressed in Nm2/kg2
The gravitational field strength, also called acceleration due to gravity, ‘g’ is given by:
Hence, g=GMr2 ——-(3)
This is the relationship between gravitational constant G and acceleration due to gravity g.
Therefore, we define gravitational field strength ‘g’ as force per unit mass. It is a vector quantity.
Newton Biography and Gravitation SLIDE SHOW:
Gravitational Potential Vg
The gravitational potential Vg at a point is the work done in taking a unit mass from infinity to that point on the surface of the earth.
∴ Vg=GMr ——-(5)
Where M is the mass of the earth and r is the radius of the earth of value 6.4 × 106m or 6400km
At any point, distance r from the centre of the earth, the gravitational potential experienced by a body of mass m is given by:
Since the potential at infinity is taken to be zero. (i.e, ∆potential =0−Vg=−Vg=−Gmr
The negative sign indicates that potential at infinity is higher than the potential close to the mass, that is, Vg decreases as r increases.
Escape Velocity V0
Consider a rocket of mass m placed at the centre of the earth’s surface O. If it is fired from that point so that it just escapes the earth’s gravitational field, it has a kinetic energy, k.e given as:
But work is done(WD) in taking this rocket to a distance R so great that the gravitational field is negligibly weak.
∴ Work done =mg×distance
This work done must be equal to the kinetic energy of the rocket at the point of take off
Work done =12mV20
∴ Work done =mg×R=mgR
but gravitational intensity g=GMR2
Work done =m(GMR2)R
Work done =GMmR
But k.e = WD
∴ V0=2GMR−−−−√ ——-(8)
Hence, V0=2gR−−−−√ ——-(8)
We thus define the escape velocity V0 as the velocity which is sufficient enough for a body to just escape the earth’s gravitational field.
Since g = 9.8m/s2, and R = 6,400,000m
EVALUATION (POST YOUR ANSWERS USING THE QUESTION BOX BELOW FOR EVALUATION AND DISCUSSION):
- State Newton’s law of gravitation.
- Define escape velocity.
- Write down the mathematical expression for :
- Gravitational force. ii. Gravitational potential.
- Calculate the gravitation force of attraction between two planets of masses 1024kg and 1027kg separated by a distance of 1020(G = 6.67 × 10-11).
- Find the gravitational force between a proton and an electron if their distance of separation is 10-10m. (mp = 1.67 × 10-27kg, me = 9.11 × 10-31kg