Question

What Is Acceleration Due To Gravity On Jupiter?

Answer 1

Hint:The universal gravitational constant is related to the attractive gravitational force between two bodies separated by a distance r. The acceleration due to gravity on earth is the acceleration experienced by anybody during free fall due to the attractive gravitational force of the earth's surface. It is not a universal constant. On earth, it is usually taken to be $$9.8m{s}^{-2}$$.

Formula used:
Gravitational Force,
$$F=GMm/r^2$$
Where G = universal gravitational constant.
M= mass of the planet (earth)
m = mass of the lighter object
r = distance between the two objects
Acceleration due to gravity,
$$g=GM/r^2$$
where M= mass of the planet (earth)
r = radius of the planet

Complete step by step solution:
Given: mass of Jupiter = 319 times the mass of earth
Radius of Jupiter = 11.2 times the radius of earth
From Newton’s law of gravitation,
Force, $$F=GMm/r^2$$----- (1)
where G= universal gravitational constant = $$6.674\times {10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2}$$
Also, from Newton’s third law of motion,
$$F=mg$$------ (2)
where g = acceleration due to gravity on earth = $$9.8m{s}^{-2}$$
Equating (1) and (2)
$$g=GM/r^2$$------ (3)

Let $$$$$$M$$, $$r$$, $$M_j$$ and $$r_j$$ be masses and radii of earth and Jupiter respectively. Then according to the question,
$$M_j=319M$$--(4) and $$r_j=11.2r$$--- (5)
Using equation (3) to calculate $$g_j$$= acceleration due to gravity on Jupiter,
$$g_j=G{M}_j/{r^2}_j$$---- (6)
Substituting equations (4) and (5) in (6), we get,
$$g_j=G\times 319M/(11.2r{)}^2$$
$$\Rightarrow g_j=2.54g$$
$$\Rightarrow g_j=2.54\times 9.8$$
$$\therefore g_j=24.58m{s}^{-2}$$

Hence the acceleration due to gravity on Jupiter is $$g_j=24.58m{s}^{-2}$$.

Note: Although acceleration due to gravity is a constant, it is not a universal constant like universal gravitational constant, G. It varies on earth with change in reference surface such as when measured on a mountain or in the depths of water bodies like seas and oceans.