The asteroid Ceres has a mass of `7*10^20` `kg` and a radius of `500` `km`. What is `g` on the surface of Ceres? How much would a `99` `kg` astronaut weigh on Ceres?

To solve this question, we'll first need to understand how we can arrive at the value of `g`.

From Newton's Universal Law of Gravitation, we have:

`F=G (Mm)/R^2`

Where,

`F` is the force of attraction on the larger object by the smaller object OR the force of attraction on the smaller object by the larger object,

`G` is the universal gravitational constant, `=6.67408 × 10^-11 m^3 kg^-1 s^-2`,

`M` is the mass of the larger object,

`m` is the mass of the smaller object,

and `R` is the distance by which their centres of mass are separated.
(For a sphere, the centre of mass lies at its geometrical centre.)

But from Newton's Second Law, we have:

`F=ma`

Substituting this in the first equation, we have

`cancelma=G (Mcancelm)/R^2`

`a=GM/R^2`

For a system consisting of an object on a planet or natural satellite,

`a` is written as `g` (acceleration due to gravity)

and `R` is equal to the radius of the planet.

Since the radius of a planet is so huge, we can usually neglect small distances from it. The image looks more like this now:

Alright, so we have reached on our final equation:

`g=GM/R^2`

It's interesting to notice that the value of `g` does not depend upon the mass of the smaller object `m`, which is why both heavy and small objects will accelerate towards a planet at equal rates. See Galileo's Leaning Tower of Pisa experiment .

Now, substituting the values we have for Ceres, we get:

`g=6.67408 × 10^-11 * (7*10^20)/(500*10^3)^2`

`g=1.868*10^-1 m/s^2`

or, `g=0.1868 m/s^2`

Now for the second part, we know that the weight of an object `(W)` is simply its mass `(m)` times the acceleration due to gravity `(g)`.

`W=mg`

Substituting the values we have,

`W=99*0.1868`

`W=18.4932N`

(You can compare this with `99*9.81=971.19N` on Earth.)

In case you want to know what a weighing scale from Earth would show in case the astronaut stepped on it on Ceres, you can divide his weight by `g_(earth)=9.81 m/s^2`,

which is `18.4932/9.81 = 1.8851 kg`

That's just what a weighing scale would show as we'd feel lighter on Ceres. The actual mass of the astronaut has not decreased!