Problem 55 - University Physics with Modern Physics

Question

An astronaut, whose mission is to go where no one has gone before, lands on a spherical planet in a distant galaxy. As she stands on the surface of the planet, she releases a small rock from rest and finds that it takes the rock 0.480 s to fall 1.90 m. If the radius of the planet is 8.60 \(\times\) 10\(^7\) m, what is the mass of the planet?

Step-by-Step Solution

Answer

The mass of the planet is approximately \(4.87 \times 10^{24}\,\text{kg}\).

Step 1: Determine the acceleration due to gravity

To find the acceleration due to gravity (\(g\)) on the planet, we use the formula for the distance covered by an object under constant acceleration: \[d = \frac{1}{2} g t^2\]Here, \(d = 1.90\,\text{m}\) and \(t = 0.480\,\text{s}\). Plugging in these values, we get:\[1.90 = \frac{1}{2}g (0.480)^2\]Solving for \(g\), we find:\[g = \frac{2 imes 1.90}{0.480^2}\approx 16.46\,\text{m/s}^2\]

Step 2: Use the formula for gravitational force

The gravitational force acting on a body is given by the equation:\[F = \frac{G M m}{r^2}\]where \(G = 6.674 \times 10^{-11}\,\text{Nm}^2/\text{kg}^2\) is the gravitational constant, \(M\) is the mass of the planet, \(m\) is the mass of the rock (which cancels out later), and \(r\) is the radius of the planet.

Step 3: Relate acceleration due to gravity to gravitational force

The acceleration due to gravity \(g\) can be expressed as:\[g = \frac{G M}{r^2}\]We know \(g = 16.46\,\text{m/s}^2\) from Step 1 and \(r = 8.60 \times 10^7\,\text{m}\). Solving for \(M\), the mass of the planet, we equate:\[16.46 = \frac{6.674 \times 10^{-11} M}{(8.60 \times 10^7)^2}\]

Step 4: Solve for the mass of the planet

Rearrange the equation from Step 3 to solve for \(M\):\[M = \frac{16.46 \times (8.60 \times 10^7)^2}{6.674 \times 10^{-11}}\]Calculate \(M\):\[M \approx 4.87 \times 10^{24}\,\text{kg}\]

Key Concepts

Acceleration Due to Gravity

Understanding acceleration due to gravity is crucial in comprehending how objects behave under the influence of gravity. On different planets, this acceleration can vary due to the mass and radius of the planet. This force is the key reason why objects fall towards the surface.
In physics, the acceleration due to gravity is often symbolized by the letter \( g \). On Earth, we experience an average \( g \) of approximately 9.81 m/s². However, this value changes on other celestial bodies. For instance, a smaller or less massive planet will exert less gravitational pull, resulting in a lower \( g \). Conversely, a more massive planet will have a higher \( g \). Understanding this variation is essential when calculating movements and trajectories on different planetary surfaces.
In the given exercise, using the formula \(d = \frac{1}{2} g t^2\), where \(d\) is distance and \(t\) is the time, helped us determine the acceleration due to gravity on the new planet. By rearranging and solving this equation, we found that \( g \) for this planet is 16.46 m/s², which is significantly higher than on Earth.

Spherical Planets

When considering celestial bodies like planets, their shape plays a role in how we calculate gravitational forces. Most planets, including the one in our exercise, are approximately spherical in shape. This assumption simplifies calculations because the distance from the planet’s center to any point on its surface (the radius) is consistent.
Knowing the radius is pivotal since it affects the calculation of gravitational force. Therefore, the formula for gravitational force includes the radius: \( F = \frac{G M m}{r^2} \), where \( r \) is the planet's radius.
In our exercise, with the given radius of 8.60 × 10 extsuperscript{7} m, we combine this with our previously calculated \( g \) to solve for the planet's mass. Due to being spherical, we apply standard spherical formulas without needing complex geometry adjustments, making calculations far more manageable.

Newton's Laws of Motion

Newton's Laws of Motion, especially his universal law of gravitation, help us understand the movement of objects and forces acting within our universe. Newton's law states that every point mass attracts every other point mass with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This principle forms the foundation of calculating gravitational force.
The formula \( F = \frac{G M m}{r^2} \) arises from this universal law. In our exercise, the astronaut determines gravitational acceleration which is crucial for deriving the planet’s mass.
Newton’s formulations allow predicting how quickly the rock falls and the net forces acting upon it. In this respect, laws enabling the understanding of free fall and gravity provide tools for broader applications, from space travel to engineering on Earth. Newton’s contribution makes it possible to calculate variables like the mass of a planet by merely observing an object's motion under gravity.