Chapter 1: Q37E (page 427)

The acceleration due to gravity at the north pole of Neptune is approximately $11.2 m / s^{2}$ . Neptune has mass $1.02 \times 10^{26} k m$ and radius $2.46 \times 10^{4} k m$ and rotates once around its axis in about 16 h.
(a) What is the gravitational force on a3.00 kgobject 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.)

Short Answer

Expert verified

The apparent weight of the Neptune is 32.7 N.

Step by step solution

Identification of the given data

The given data can be listed below as,

The gravity is, $= 11.2 m / s^{2}$
The mass is, $= 1.02 \times 10^{6} k m$
The radius is, $= 2.46 \times 10^{4 k m} 2.46 \times 10^{4 k m}$
The surface is gaseous, not solid, so it is impossible to stand on it.

The gravitational force on a object at north pole of Neptune

Gravitational force,

F = mg

Here, m is the mass of the Neptune, g is the gravity of the Neptune.

Substitute all the values in the above equation.

$\begin{array}{l} F = m g \\ F = 3 \times 11.2 = 33.6 N \end{array}$

The apparent weight of the object at Neptune’s equator

Apparent weight at the equator = m(g-a)

a = centripetal acceleration at equator = $R ω^{2}$

$16 hr = 57600 s ω = \frac{2 π}{T} = \frac{2 π}{57600 s} = 1.1 \times 10^{- 4} rad / \sec a = {Rω}^{2} = 0.0246 \times 10^{6} \times {(1.1 \times 10^{- 4})}^{2} = 0.297 \approx 0.30 m / s^{2}$

Apparent weight = m(g-a)

$\begin{array}{l} = 3 a (11.2 - 0.3) \\ = 32.7 N \end{array}$

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Short Answer

Step by step solution

Identification of the given data

The gravitational force on a object at north pole of Neptune

The apparent weight of the object at Neptune’s equator

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