Question

A simple instrument for measuring the acceleration of gravity employs a linear spring from which a mass is suspended. At a location on earth where the acceleration of gravity is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\), the spring extends \(0.739 \mathrm{~cm}\). If the spring extends \(0.116\) in. when the instrument is on Mars, what is the Martian acceleration of gravity? How much would the spring extend on the moon, where \(g=1.67 \mathrm{~m} / \mathrm{s}^{2}\) ?

Short answer

The Martian gravity is 3.91 m/s², and the spring would extend 0.12588 cm on the Moon.

Step-by-step solution

Convert units

First, convert all measurements to consistent units. Since the given gravity on Earth is in meters per second squared (m/s²), convert the spring extension measurements to meters. For Earth: 0.739 cm to meters: \(0.739~cm = 0.739~cm~\times~\frac{1~m}{100~cm} = 0.00739~m\) For Mars: 0.116 inches to meters: \(0.116~in = 0.116~in~\times~\frac{2.54~cm}{1~in}~\times~\frac{1~m}{100~cm} = 0.0029464~m\)

Apply Hooke's Law

Hooke's Law relates the force exerted by a spring, the spring constant \(k\), and the displacement \(x\) by \(F = kx\). On Earth, the force is the weight of the mass \(( F = mg )\), where \( m \) is the mass and \( g \) is the acceleration due to gravity. \( mg_{Earth} = k x_{Earth} \) Rearranging, \( k = \frac{ mg_{Earth}}{x_{Earth}} \)

Calculate Spring Constant

Substitute known values from Earth \( g_{Earth}=9.81 \, m/s^2 \), and \( x_{Earth} = 0.00739 \, m \): \( k = \frac{ mg_{Earth}}{x_{Earth}} = \frac{ m \times 9.81 }{0.00739} \, N/m \)

Determine acceleration on Mars

Knowing \( k \) is constant, apply it to calculate Martian gravity. \( mg_{Mars} = k x_{Mars} \) Rearranging, \( g_{Mars} = \frac{k x_{Mars}}{ m } \) Substitute \( k \) and \(x_{Mars} = 0.0029464 \): \( g_{Mars} = \frac{ 9.81 \times 0.0029464}{0.00739} = 3.91 \, m/s^2 \)

Determine extension on Moon

Use the same constant \( k \) to find the extension on the Moon. From Hooke's Law: \( mg_{Moon} = k x_{Moon} \) Rearranging to solve for \( x_{Moon} \): \( x_{Moon} = \frac{ m \times g_{Moon}}{k} \) Substitute \( k \) and \(g_{Moon} = 1.67 \, m/s^2 \): \( x_{Moon} = \frac{ 1.67 \times 0.00739}{9.81} = 0.0012588 \, m = 0.12588 \, cm \)

Key concepts

Hooke's Law

Hooke's Law is fundamental for understanding the behavior of springs. It states that the force exerted by a spring is directly proportional to the displacement or extension of the spring. This can be expressed mathematically as:
\( F = kx \)
Here, \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement or extension of the spring.

When you hang a mass from a spring, the force due to gravity on the mass (its weight) causes the spring to extend. The weight, \( mg \), where \( m \) is mass and \( g \) is gravitational acceleration, is balanced by the spring's force: \( mg = kx \). This relationship helps us determine the spring constant and understand how the spring will behave under different gravitational conditions.

Unit Conversion

Unit conversion is crucial for ensuring that all quantities are measured in consistent units, especially when applying physical laws like Hooke's Law.
In the exercise, we need to convert lengths to meters.

• Convert 0.739 cm to meters: \(0.739 \times \frac{1}{100} = 0.00739 \text{ m} \)
• Convert 0.116 inches to meters: \(0.116 \times 2.54 \times \frac{1}{100} = 0.0029464 \text{ m} \)

By converting both measurements to meters, we ensure that our calculations are consistent and the results are accurate.

Spring Constant

The spring constant \( k \) is a measure of a spring's stiffness and is defined as the ratio of the force applied to the displacement it causes: \( k = \frac{F}{x} = \frac{mg}{x} \).
In the exercise, by substituting the known values on Earth, we can calculate the spring constant:

If \( g_{Earth} = 9.81 \text{ m/s^2} \) and \( x_{Earth} = 0.00739 \text{ m} \), the spring constant is:
\[ k = \frac{m \times 9.81}{0.00739} \text{ N/m} \].
This constant remains the same regardless of the location, and it helps us calculate the gravitational acceleration on other planets or moons when the displacement is known.

Gravity on Earth

Gravity on Earth is approximately \(9.81 \text{ m/s}^2 \). This familiar acceleration due to gravity is what we used in the initial calculation to determine the spring constant. Using Hooke's Law, we can rearrange the formula to find \( k \):

\[ k = \frac{m \times 9.81}{0.00739} \text{ N/m} \].
Earth's gravitational pull is essential for understanding the baseline measurements which we use to compare with other celestial bodies like Mars and the Moon.

Gravity on Mars

The acceleration due to gravity on Mars differs significantly from Earth. To find it, we use the same spring constant obtained from Earth but with the Martian displacement:
\[ mg_{Mars} = k x_{Mars} \]
By rearranging the formula, we get:
\[ g_{Mars} = \frac{k x_{Mars}}{m} \].
Substituting the values, \[ g_{Mars} = \frac{9.81 \times 0.0029464}{0.00739} = 3.91 \text{ m/s}^2 \], we find that Mars has less gravitational pull than Earth, impacting how the spring extends.

Gravity on Moon

Gravity on the Moon is even weaker than on Mars or Earth, at approximately \(1.67 \text{ m/s}^2 \). Using the same spring constant, we can determine the extension of the spring under lunar gravity:

\[ mg_{Moon} = k x_{Moon} \]
Solving for \( x_{Moon} \), we get:
\[ x_{Moon} = \frac{m \times 1.67}{k} \].
Substituting our known values,
\[ x_{Moon} = \frac{1.67 \times 0.00739}{9.81} = 0.0012588 \text{ m} = 0.12588 \text{ cm} \].
Understanding lunar gravity helps illustrate how gravitational forces vary across different celestial bodies.