Question

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\(^{26}\) kg and radius 2.46 \(\times\) 10\(^4\) km 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.)

Short answer

(a) 33.6 N (gravitational force at north pole); (b) 33.51 N (apparent weight at equator).

Step-by-step solution

Gravitational Force at the North Pole

The gravitational force exerted on an object is given by the formula: \[ F_g = m \cdot g \]where \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity. Here, \( m = 3.00 \text{ kg} \) and \( g = 11.2 \text{ m/s}^2 \). Substituting these values, we get:\[ F_g = 3.00 \cdot 11.2 = 33.6 \text{ N} \] So, the gravitational force on the object at the north pole is 33.6 N.

Calculating the Centripetal Force at the Equator

At the equator, the object experiences a centrifugal effect due to Neptune's rotation. To find the centripetal force, use the formula:\[ F_c = m \cdot \omega^2 \cdot r \]where \( \omega \) is the angular velocity and \( r \) is the radius.First, find \( \omega \) using:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the rotational period (16 h converted to seconds: \( 16 \cdot 3600 = 57600 \text{ s} \)). Thus:\[ \omega = \frac{2\pi}{57600} \approx 1.09 \times 10^{-4} \text{ rad/s} \]Now, substitute \( \omega \) and \( r = 2.46 \times 10^7 \text{ m} \) (converted from km to m):\[ F_c = 3.00 \cdot \left(1.09 \times 10^{-4}\right)^2 \cdot 2.46 \times 10^7 \approx 0.0877 \text{ N} \]

Apparent Weight at the Equator

The apparent weight is the actual gravitational force minus the centripetal force due to rotation. This is given by:\[ W_{apparent} = F_g - F_c \]Using the values from the previous steps:\[ W_{apparent} = 33.6 - 0.0877 \approx 33.5123 \text{ N} \]So, the apparent weight of the object at Neptune's equator is approximately 33.5123 N.

Key concepts

Centripetal Force

Centripetal force is a crucial concept in understanding motion, especially in scenarios involving curves or rotations. It is the force required to keep an object moving in a circular path and is directed toward the center of the circle. Imagine swinging a ball attached to a string in a circle; the tension in the string is a classic example of centripetal force because it pulls the ball toward the center, preventing it from flying away.

In the context of planetary rotation, such as that of Neptune, the object at the equator experiences a centripetal force due to the planet's rotation. This force can be calculated using the formula:

• \( F_c = m \cdot \omega^2 \cdot r \)

where \( m \) is the mass of the object, \( \omega \) is the angular velocity, and \( r \) is the radius of rotation.

Angular velocity \( \omega \) is another important aspect of centripetal force. It represents how fast an object rotates about its axis. For Neptune, it can be determined by the formula:

• \( \omega = \frac{2\pi}{T} \)

where \( T \) is the rotational period of the planet. The centripetal force acts as a counter to the gravitational force, affecting the apparent weight of objects there.

Apparent Weight

Apparent weight is the perceived weight of an object, which can differ from the actual weight due to motion effects, such as centripetal acceleration. Apparent weight is especially prominent in non-inertial reference frames where acceleration occurs. This concept becomes important when considering how objects behave on rotating planets like Neptune.

For an object at Neptune's equator, the effect of the planet's rotation slightly reduces its apparent weight compared to its true gravitational force. The force exerted on an object here must account for the subtractive effect of the centripetal force experienced due to the rotation. This can be calculated using:

• \( W_{apparent} = F_g - F_c \)

where \( F_g \) is the actual gravitational force and \( F_c \) is the centripetal force.

Using these calculations, students can understand why apparent weight differs from actual weight and how dynamics like rotation influence this phenomenon.

Neptune's Gravity

Neptune, the eighth planet from the Sun, possesses a distinct gravity that is slightly stronger than Earth's, despite its gaseous nature. With a gravitational acceleration of approximately \(11.2 \text{ m/s}^2\), the force it exerts is significantly influenced by its massive size and considerable radius.

Gravity on Neptune acts similarly to gravity on other celestial bodies; it attracts objects toward the planet's center, contributing to their weight. Neptune's huge mass, \(1.02 \times 10^{26} \text{ kg}\), plays a central role in determining this gravitational pull. When considering other planets, it is essential to understand how mass and radius affect gravity according to the equation:

• \( g = \frac{G \cdot M}{r^2} \)

where \( g \) is the acceleration due to gravity, \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( r \) is the radius.

For students, grasping the concept of Neptune's gravity provides insight into how planetary characteristics like mass and rotational speed influence gravitational effects. This understanding is crucial when exploring dynamics on any planetary body.