Suppose all the mass of the earth were compacted into a small spherical ball. part a what radius must the sphere have so that the acceleration due to gravity at the earth's new surface was equal to the acceleration due to gravity at the surface of the moon?

0.4 times the radius of moon

Explanation:

gravity on moon is equal to the one sixth of gravity on earth.

g' = g / 6

where, g' is the gravity on moon and g be the gravity on earth.

As the earth shrinks, the mass of earth remains same.

The acceleration due to gravity is inversely proportional to the square of radius of planet.

g' ∝ 1/R'²   (1)

Where, R' is the radius of moon.

g ∝ 1/R²       (2)

Where, R be the radius of earth.

Divide equation (1) by (2)

g / g' = R'² / R²

Put g' = g / 6

6 = R'² / R²

2.5 = R' / R

R = R' / 2.5 = 0.4 R'

Thus, the radius of earth should be 0.4 times the radius of moon.

a skateboard rolling across the street - apex

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures