Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

Solutions 209

to the force of pressure of the atmosphere, i.e. the
pressure on the surface of Mars is pm = mmgm/
(43-cRk). Similarly, for the corresponding param-

eters on the Earth, we obtain PE = mEgE/(4a/q).

The ratio of the masses of the Martian and the

Earth's atmospheres is

mm pm •4n/itigE

mE pE •4nRtgm

Considering that MM = (4/3)nnpm (a similar

expression can be obtained for the Earth) and sub-

stituting the given quantities, we get

mm pm Rm PE

= — — fsz 3.4 X 10-3.

mE PE RE PM

It should be noted that we assumed in fact that

the atmosphere is near the surface of a planet.

This is really so since the height of the atmosphere

is much smaller than the radius of a planet (e.g.

at an altitude of 10 km above the surface of the

Earth it is impossible to breathe, and the radius

of the Earth is RE f_se. 6400 km!).