Thermal, Quantum, and Nuclear Physics
Gravitation
Gravitational field strength at a point is defined as the gravitational force per unit mass at that point.
Newton's law of gravitation:
The (mutual) gravitational force F between two point masses M and m separated by a distance r is given by F = | GMm | (where G: Universal gravitational constant) | | r2 | | or, the gravitational force of between two point masses is proportional to the product of their masses & inversely proportional to the square of their separation.
Gravitational field strength at a point is the gravitational force per unit mass at that point. It is a vector and its S.I. unit is N kg-1.
By definition, g = F / m
By Newton Law of Gravitation, F = GMm / r2
Combining, magnitude of g = GM / r2
Therefore g = GM / r2, M = Mass of object “creating” the field
Example 1:
Assuming that the Earth is a uniform sphere of radius 6.4 x 106 m and mass 6.0 x 1024 kg, find the gravitational field strength g at a point:
(a) on the surface, g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / (6.4 x 106)2 = 9.77ms-2
(b) at height 0.50 times the radius of above the Earth's surface. g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / ( (1.5 × 6.4 x 106)2 = 4.34ms-2
Example 2:
The acceleration due to gravity at the Earth's surface is 9.80ms-2. Calculate the acceleration due to gravity on a planet which has the same density but twice the radius of Earth. g = GM / r2 gP / gE = MPrE2 / MErP2 = (4/3) π rP3rE2ρP / (4/3) π rE3rP2ρE = rP / rE = 2
Hence gP = 2 x 9.81 = 19.6ms-2
Assuming that Earth is a uniform sphere of mass M. The magnitude of the gravitational force from Earth on a particle of mass m, located outside Earth a distance r from the centre of the Earth is F = GMm / r2. When a particle is released, it will fall towards the centre of the Earth, as a result of the gravitational force with an acceleration ag.
FG = mag ag = GM / r2
Hence ag = g
Thus gravitational field strength g is also numerically equal to the acceleration of free fall.
Gravitational field strength at a point is defined as the gravitational force per unit mass at that point.
Newton's law of gravitation:
The (mutual) gravitational force F between two point masses M and m separated by a distance r is given by F = | GMm | (where G: Universal gravitational constant) | | r2 | | or, the gravitational force of between two point masses is proportional to the product of their masses & inversely proportional to the square of their separation.
Gravitational field strength at a point is the gravitational force per unit mass at that point. It is a vector and its S.I. unit is N kg-1.
By definition, g = F / m
By Newton Law of Gravitation, F = GMm / r2
Combining, magnitude of g = GM / r2
Therefore g = GM / r2, M = Mass of object “creating” the field
Example 1:
Assuming that the Earth is a uniform sphere of radius 6.4 x 106 m and mass 6.0 x 1024 kg, find the gravitational field strength g at a point:
(a) on the surface, g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / (6.4 x 106)2 = 9.77ms-2
(b) at height 0.50 times the radius of above the Earth's surface. g = GM / r2 = (6.67 × 10-11)(6.0 x 1024) / ( (1.5 × 6.4 x 106)2 = 4.34ms-2
Example 2:
The acceleration due to gravity at the Earth's surface is 9.80ms-2. Calculate the acceleration due to gravity on a planet which has the same density but twice the radius of Earth. g = GM / r2 gP / gE = MPrE2 / MErP2 = (4/3) π rP3rE2ρP / (4/3) π rE3rP2ρE = rP / rE = 2
Hence gP = 2 x 9.81 = 19.6ms-2
Assuming that Earth is a uniform sphere of mass M. The magnitude of the gravitational force from Earth on a particle of mass m, located outside Earth a distance r from the centre of the Earth is F = GMm / r2. When a particle is released, it will fall towards the centre of the Earth, as a result of the gravitational force with an acceleration ag.
FG = mag ag = GM / r2
Hence ag = g
Thus gravitational field strength g is also numerically equal to the acceleration of free fall.