The free-fall acceleration on the surface of mercury is about three eighths that on the surface of the earth. the radius of mercury is about 0.375 re (re = earth's radius = 6.4 106 m). find the ratio of their average densities, ? mercury/? earth.

1

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

= Mass of Earth

= Mass of Mercury

= Radius of Earth

= Radius of Mercury =

Mass is given by

Acceleration due to gravity on Earth

Acceleration due to gravity on Mercury

According to the question

So,

Hence, the average density ratio of Mercury to Earth is 1

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west