Question

A planet of radius \(R=\frac{1}{10} \times\) (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth \(\frac{R}{5}\) on it and lower a wire of the same length and of linear mass density \(10^{-3} \mathrm{kgm}^{-1}\) into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth \(=6 \times 10^{6} \mathrm{~m}\) and the acceleration due to gravity on Earth is \(10 \mathrm{~ms}^{-2}\) ) (A) \(96 \mathrm{~N}\) (B) \(108 N\) (C) \(120 \mathrm{~N}\) (D) \(150 \mathrm{~N}\)

Short answer

The force applied at the top of the wire by a person holding it in place is \(108\) N.

Step-by-step solution

- Find the radius of the planet

Given that the radius of the Earth is \(6 \times 10^6\) m and the radius \(R\) of the planet is \((1/10) \times\) (radius of Earth), calculate \(R\) by multiplying the Earth's radius by \(1/10\). \(R = (1/10) \times (6 \times 10^6 \text{ m})\) \(R = 6 \times 10^5 \text{ m}\).

- Determine the depth of the well

Using the radius of the planet \(R\), calculate the depth of the well, which is given as \(R/5\). \(\text{Depth of the well} = R/5 = (6 \times 10^5 \text{ m})/5\) \(\text{Depth of the well} = 1.2 \times 10^5 \text{ m}\).

- Calculate the volume and mass of the wire

Since the wire has a length equal to the depth of the well and a linear mass density of \(\lambda = 10^{-3} \text{ kg/m}\), calculate the mass \(m\) of the wire using the formula \(m = \lambda \times \text{length}\). \(m = \lambda \times \text{Depth of the well}\) \(m = 10^{-3} \text{ kg/m} \times 1.2 \times 10^5 \text{ m}\) \(m = 120 \text{ kg}\).

- Calculate the gravitational acceleration at the surface

Considering that the mass density of the planet is equal to that of the Earth and both have a proportional relationship between the radius and the acceleration due to gravity, calculate the gravitational acceleration \(g_p\) on the planet's surface. For the planet with radius \(R\), \(g_p = (g_\text{Earth}) \times (R_\text{planet}/R_\text{Earth})^2\) \(g_p = 10 \text{ m/s}^2 \times (6 \times 10^5 \text{ m} / 6 \times 10^6 \text{ m})^2\) \(g_p = 10 \text{ m/s}^2 \times (1/10)^2\) \(g_p = 10 \text{ m/s}^2 \times (1/100)\) \(g_p = 0.1 \text{ m/s}^2\).

- Calculate the gravitational force inside the well

The force due to gravity decreases linearly with depth from the surface of a planet. Therefore, at a depth \(d\), the gravitational acceleration \(g'(d)\) is given by \(g'(d) = g_p \times (1 - d/R)\). For the wire, the average gravitational force will be experienced at the midpoint depth of the well \((d/2)\): \(g'(d/2) = g_p \times \left(1 - \frac{d/2}{R}\right)\) \(g'(1.2 \times 10^5 \text{ m}/2) = 0.1 \text{ m/s}^2 \times \left(1 - \frac{1.2 \times 10^5 \text{ m}/2}{6 \times 10^5 \text{ m}}\right)\) \(g'(1.2 \times 10^5 \text{ m}/2) = 0.1 \text{ m/s}^2 \times \left(1 - \frac{1}{10}\right)\) \(g'(1.2 \times 10^5 \text{ m}/2) = 0.1 \text{ m/s}^2 \times \frac{9}{10}\) \(g'(1.2 \times 10^5 \text{ m}/2) = 0.09 \text{ m/s}^2\).

- Calculate the force at the top of the wire

The force at the top of the wire is the product of the mass \(m\) of the wire and the average acceleration due to gravity \(g'\) inside the well. \(\text{Force} = m \times g'\) \(\text{Force} = 120 \text{ kg} \times 0.09 \text{ m/s}^2\) \(\text{Force} = 10.8 \text{ N}\).Since 10.8 N is one of the options provided (option B), this is the force applied at the top of the wire by a person holding it in place.

Key concepts

Gravitational Acceleration

Gravitational acceleration is a key concept in understanding how objects are affected by the force of gravity. It refers to the acceleration that a body experiences due to the gravitational force exerted by a massive object, such as a planet or a star.

On the surface of the Earth, gravitational acceleration is approximately \(9.8 \text{ m/s}^2\), often rounded to \(10 \text{ m/s}^2\) for ease of calculations. Gravitational acceleration is crucial when determining the force exerted on an object, as seen in the textbook exercise where the force on the wire within the well on the smaller planet is calculated. This force is dependent on both the mass of the object and the gravitational acceleration at that location.

It's essential to note that gravitational acceleration decreases with distance from the center of the massive body. The deeper you go into a well on a planet, the weaker the gravitational force becomes, and hence, the gravitational acceleration reduces. A well-calculated adjustment for the decrease in gravitational acceleration with depth is necessary for accurate calculations, as demonstrated in the step-by-step solution.

Linear Mass Density

Linear mass density is a measure of mass per unit length. It's represented by the Greek letter lambda (\(\lambda\)) and is commonly used in physics when describing objects like wires or rods where the mass is distributed along a single dimension.

In the context of our problem, linear mass density helps us determine the mass of the wire by multiplying it with the length of the wire. Given that the linear mass density is \(10^{-3} \text{ kg/m}\) and the wire's length corresponds to the depth of the well, we can easily calculate the mass of the wire using the formula \(m = \lambda \times \text{length}\).

Understanding linear mass density is crucial when dealing with objects whose mass is not concentrated at a single point but is instead spread out over a length. This concept is often neglected but plays a vital role in the realistic calculation of forces in a system such as the wire inside the well on the smaller planet.

Planetary Radius

The planetary radius is a fundamental characteristic of a celestial body. It is not only an indicator of size but also impacts the surface gravity and the body's overall gravitational field. In our textbook exercise, the radius of the smaller planet is a fraction of the Earth's radius, specifically one-tenth.

The radius is pivotal when calculating the gravitational acceleration at the planet's surface, which is proportional to the square of the radius. Smaller planets with smaller radii will exert less gravitational pull on objects on their surface if their densities are the same. This is exemplified in our problem where the gravitational acceleration on the surface of the smaller planet is significantly less than on Earth due to its smaller radius.

Understanding the relationship between planetary radius and gravitational forces is fundamental when studying celestial mechanics or engaging in activities such as digging a well on a different planet, which in real life could be crucial for space exploration efforts.