Question

If a small asteroid named ‘B612’ has a radius of only 20 m and mass of

Answer 1

Acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity.

It is a vector quantity and it's SI unit is m/s².

It is usually denoted by the symbol g.

Formula of Acceleration due to Gravity:

$\boxed{ \bf{g = \dfrac{GM}{r^2}}}$

G → Universal Gravitational Constant $\sf (6.67 \times 10^{-11} \ Nm^2/kg^2)$

M → Mass of the Asteroid (1 × 10⁴ kg)

R → Radius of the Asteroid (20 m)

By substituting values in the equation, we get:

$\rm \longrightarrow g = \dfrac{6.67 \times {10}^{ - 11} \times 1 \times {10}^{4} }{20^2} \\ \\ \rm \longrightarrow g = \dfrac{6.67 \times {10}^{ - 11 + 4} }{400} \\ \\ \rm \longrightarrow g = 0.016675 \times {10}^{ - 7} \\ \\ \rm \longrightarrow g = 1.6675 \times {10}^{ -9} \: m {s}^{ - 1}$

$\therefore$ Acceleration due to gravity on the surface of small asteroid named ‘B612’ = $\sf 1.6675 \times 10^{-9} \ m/s$