Solution 2.10

From Mechanics

The Law of gravitation applied to a particle on Mars gives,

F=d2Gm1m2

where F is the force on a particle on the surface of Mars, m1 is the mass of Mars, m2 is the mass of the particle and d is the radius of Mars.

As F=m2a where a is the acceleration of the particle on Mars,

a=d2Gm1

The radius of Mars must be expressed in SI units.

d=3.4106 m

and as G=6.6710−11 kg-1m3s-2 we get

a=3.4106m26.6710−11 kg-1m3s-2 6.421023kg=3.426.676.42 ms−2=3.7 ms−2