To fix this question, we"ll very first need come understand just how we deserve to arrive in ~ the value of #g#.

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From Newton"s Universal law of Gravitation, us have:

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

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Where,

#F# is the pressure of attraction on the larger object by the smaller object OR the force of attraction top top the smaller sized object by the larger object,

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

#M# is the massive of the larger object,

#m# is the mass of the smaller sized object,

and #R# is the distance whereby their centres that mass space separated.(For a sphere, the center of massive lies at its geometrical centre.)

But from Newton"s second Law, we have:

#F=ma#

Substituting this in the first equation, us have

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

#a=GM/R^2#

For a device consisting of things on a world or organic satellite,

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

and #R# is equal to the radius the the planet.

Since the radius that a earth is for this reason huge, we have the right to usually neglect small distances from it. The photo looks an ext like this now:

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Alright, for this reason we have reached on our final equation:

#g=GM/R^2#

It"s exciting to notice that the worth of #g# walk not depend upon the massive of the smaller sized object #m#, i m sorry is why both hefty and little objects will accelerate in the direction of a planet at same rates.

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View Galileo"s Leaning Tower the Pisa experiment .

Now, substituting the worths we have actually 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 because that the 2nd part, we understand that the load of things #(W)# is merely its fixed #(m)# time the acceleration as result of gravity #(g)#.

#W=mg#

Substituting the worths we have,

#W=99*0.1868#

#W=18.4932N#

(You can compare this with #99*9.81=971.19N# ~ above Earth.)

In case you desire to know what a weighing scale from earth would show in situation the astronaut stepped on the on Ceres, you have the right to divide his weight by #g_(earth)=9.81 m/s^2#,

which is #18.4932/9.81 = 1.8851 kg#

That"s simply what a weighing range would present as we"d feel lighter ~ above Ceres. The actual massive of the astronaut has not decreased!