Two planets A𝐴 and B𝐵 of radii R𝑅 and 1.5R1.5𝑅 have densities ρ𝜌 and ρ/2𝜌/2 respectively. The ratio of acceleration due to gravity at the surface of B𝐵 to A is
Question
Two planets A𝐴 and B𝐵 of radii R𝑅 and 1.5R1.5𝑅 have densities ρ𝜌 and ρ/2𝜌/2 respectively. The ratio of acceleration due to gravity at the surface of B𝐵 to A is
Solution
The acceleration due to gravity on the surface of a planet is given by the formula:
g = GM/R^2
where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.
The mass of a planet can be found by the formula:
M = ρV
where ρ is the density of the planet and V is the volume of the planet.
The volume of a planet can be found by the formula:
V = 4/3πR^3
So, the mass of planet A is:
M_A = ρ * 4/3πR^3
and the mass of planet B is:
M_B = (ρ/2) * 4/3π(1.5R)^3
The acceleration due to gravity on the surface of planet A is:
g_A = G * M_A / R^2 = G * ρ * 4/3πR^3 / R^2 = 4/3 GρπR
The acceleration due to gravity on the surface of planet B is:
g_B = G * M_B / (1.5R)^2 = G * (ρ/2) * 4/3π(1.5R)^3 / (1.5R)^2 = 2/3 GρπR
So, the ratio of the acceleration due to gravity at the surface of planet B to that of planet A is:
g_B / g_A = (2/3 GρπR) / (4/3 GρπR) = 1/2
Therefore, the acceleration due to gravity at the surface of planet B is half that of planet A.
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