Question

What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

Answer 1

Answer:

Explanation:

Acceleration of gravity on the surface of

Neptune is given

g' = G M/R^2,

where M is the planet's mass.

M g' will be the apparent weight registed at the poles.

G is the universal constant of gravity

The "apparent weight" at the equator will be

reduced from M g' by the centripetal force M

R w^2, where w is the rotational angular

velocity in radians/s), which you can deduce

from the 16 hour length of a day.

Answer 2

Answer: (a) Fg= Weight of object = 33.6N

(b) Apparent weight of object = 32.7N

Explanation:

Here is the complete question:

The acceleration due to gravity at the north pole of Neptune is approximately 11.2m/s2. Neptune has mass 1.02×1026kg and radius 2.46×104km and rotates once around its axis in about 16 h. (a) What is the gravitational force on a 3.00-kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune’s equator? (Note that Neptune’s “surface” is gaseous, not solid, so it is impossible to stand on it.)

Explanation:

Please see the attachments below