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!