Differentiate between acceleration due to gravity and universal gravitational constant. Derive a relation between ‘g' and ‘g'. (b) state universal law of gravitation.
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Gravitational force is given by
F = GMm/r^2
Here F = ma = mg
Therefore,
mg = GMm/r^2
Divide both sides by m
g = GM / r^2
In above g and G relation
g = acceleration due to gravity of planet
G = universal gravitational constant
M = mass of planet
r = radius of planet
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Answer:
ANSWER :- (a)
'G'
(a) It is the universal gravitational constant
(b) Value of G = 6.67 × 10^-11 Nm^2 / kg^2 remain constant at very point.
(c) It is equal to force between two points maases separated by unit distance.
(d) It has an extremely small Value.
'g'
(a) It is acceleration produced due to gravity.
(b) It value changes from point to point.
(c) It is equal to acceleration experience by a body of any mass.
(d) It has relatively large magnitude.
Acceleration due to gravity :-
(i) The gravitational force exerted by earth upon an object is called it's gravity.
(ii) The acceleration of a body near the surface of the earth due to its gravity is called acceleration due to gravity (g).
By Newton's laws of gravitational,
F = GMm / r^2 ...(i)
Also, by Newton's second law,
F = ma
i.e. F = mg ...(ii)
Equating R.H.S. of (i) and (ii),
mg = GMm / r^2
or
g = GM / r^2
(On surface of earth, g = 9.8 ms^-2)
Therefore,
if, M = mass of earth
m = mass of object
r = distance between centers of earth and object
(a = acceleration due to gravity = g)
ANSWER :- (b)
Universal law of gravitation :-
It states that the gravitational force exerted between any two objects of mass 'm^1' and 'm^ 2)' whose centre are 'r' units part, is
(i) directly proportional to the product of mass, i.e. F = m^1 m^2.
(ii) inversely proportional to square of distance between their centre,
i.e. F = 1 / r^ 2
Thus, F = m^1m m^2 / r^2
or F = G m^1 m^2 / r^2
where G is the constant of proportionality called Universal Gravitational Constant.
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