Question

If an angel visits an asteroid called B 612 which has a radius of $20{\text{ }}m$ and mass of $104{\text{ }}kg$ , what will be the acceleration due to gravity in B 612?

Answer 1

In order to solve the given question, to find out the acceleration due to gravity we will simply apply the formula of acceleration due to gravity. Now, let us understand the concept in brief before going through the problem. Acceleration due to gravity is the acceleration gained by an object due to gravitational force. $m \cdot {s^{ - 2}}$ is the SI unit for it. Because it has both magnitude and direction, it is a vector quantity. The letter $g$ stands for gravitational acceleration.

Complete answer:
We will use a common, gravity equation to solve the given problem, which gives the acceleration due to gravity, g, here on the Earth's surface:
$g = \dfrac{{GM}}{{{r^2}}}$
Where,
$M =$ is the mass of the Earth,
$r =$ the radius of the Earth (or distance between the center of the Earth and you, standing on its surface), and
$G =$ is the gravitational constant.
Now, putting the given values in the above equation we will get our desired answer
$g = \dfrac{{GM}}{{{r^2}}}$
$G =$ Universal Constant
$M =$ Mass of asteroid $= 104{\text{ }}kg$
$R =$ Radius of Asteroid $= 20{\text{ }}m$
Gravitational acceleration in Asteroid B 612
$= \dfrac{{6.67 \times {{10}^{ - 11}} \times 104}}{{{{\left( {20} \right)}^2}}}$
$= 1.7342 \times {10^{ - 11}}m{s^{ - 1}}$
The acceleration due to gravity in B 612 is $1.7342 \times {10^{ - 11}}m{s^{ - 1}}$

Note:
Some important points regarding acceleration due to gravity which should be kept in mind are:
i. The acceleration due to gravity is less for an object placed at a height h than for one placed on the surface.
ii. The value of acceleration due to gravity $\left( g \right)$ decreases as depth increases.
iii. At the poles, the value of $\left( g \right)$ is higher, while at the equator, it is lower.