Problem 27 - College Physics

Question

Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from \(5.00 \mathrm{~m}\) above the ground and measure that it hits the ground 0.811 s later. (a) What is the acceleration of gravity near the surface of this planet? (b) Assuming that the planet has the same density as that of earth \(\left(5500 \mathrm{~kg} / \mathrm{m}^{3}\right),\) what is the radius of the planet?

Step-by-Step Solution

Answer

(a) The gravitational acceleration, \( g \), is approximately \( 15.21 \, \text{m/s}^2 \). (b) The planet's radius is approximately \( 5.18 \times 10^6 \, \text{m} \).

Step 1: Identify the given values

We need to determine the acceleration due to gravity on a planet by using the fact that a wrench is released from a height. The given values are:Initial height, \( h = 5.00 \, \text{m} \)Time taken to hit the ground, \( t = 0.811 \, \text{s} \).

Step 2: Apply kinematic equation for free fall

We use the kinematic equation: \[h = \frac{1}{2}gt^2\]where \( h \) is the height, \( g \) is the acceleration due to gravity, and \( t \) is the time taken. Plug in the given values:\[5.00 = \frac{1}{2}g(0.811)^2\]This equation will help solve for \( g \).

Step 3: Solve for acceleration due to gravity, g

Rearranging the equation from Step 2:\[5.00 = \frac{1}{2}g(0.811)^2\]\[g = \frac{2 \times 5.00}{(0.811)^2}\]Calculate \( g \):\[g = \frac{10.00}{0.657321} \approx 15.21 \, \text{m/s}^2\]

Step 4: Use planet density to find radius

Assuming the density of the planet is the same as Earth's, \( \rho = 5500 \, \text{kg/m}^3 \). The planet's radius, \( R \), can be found using the equation for gravitational acceleration:\[g = \frac{G M}{R^2}\]where \( M \) is the mass of the planet, and \( G \) is the gravitational constant, \( G = 6.674 \times 10^{-11} \, \text{m}^3/\text{kg/s}^2 \).

Step 5: Find mass of the planet using volume and density

Express the mass, \( M \), in terms of volume and density:\[M = \frac{4}{3} \pi R^3 \times 5500 \]Substitute \( M \) into the equation from Step 4 and solve for \( R \).

Step 6: Solve for the radius, R

The equation becomes:\[g = \frac{G \cdot \frac{4}{3} \pi R^3 \cdot 5500}{R^2}\]Simplify to solve for \( R \):\[g = \frac{4}{3} \pi G \cdot R \cdot 5500\]Substitute for \( g = 15.21 \, \text{m/s}^2 \), and solve:\[R = \left(\frac{15.21}{\frac{4}{3} \pi \cdot G \cdot 5500}\right)\]Calculate \( R \approx 5.18 \times 10^6 \, \text{m}\).

Key Concepts

Gravitational Acceleration

Gravitational acceleration is the rate at which an object speeds up as it falls freely toward a massive body, like a planet. In this scenario, we are curious about how fast objects fall on an unknown planet. The key formula used is derived from kinematics for free fall:

• Initial height: \( h = 5.00 \, \text{m} \)
• Time to fall: \( t = 0.811 \, \text{s} \)

The gravitational acceleration, \( g \), is calculated using the formula \(h = \frac{1}{2}gt^2\).This tells us how high an object will be at any given time period as it falls. Rearranging gives us\(g = \frac{2h}{t^2}\).Plugging in our values:\[g = \frac{2 \times 5.00}{(0.811)^2} \approx 15.21 \, \text{m/s}^2\]This means that on this alien planet, objects accelerate faster than they do on Earth, where \( g \approx 9.81 \, \text{m/s}^2 \). It's a key clue about the planet's characteristics.

Free Fall

When we talk about free fall, we're describing an object moving solely under the influence of gravity, with no other forces acting like air resistance. On the unknown planet, the wrench we dropped behaved as a free-falling object. Because the equation \( h = \frac{1}{2}gt^2 \) only includes gravity, it helps simplify our calculation for \( g \). We ignore other forces for simplicity in basic physics problems like this one.
The wrench starts from rest and accelerates downwards as it falls. The time it takes to hit the ground tells us about the gravitational pull.
This concept of free fall is crucial not just for planets in our solar system but for understanding how gravity works across the universe.

Planetary Density

Density is a measure of how much mass is contained in a given volume. The exercise simplifies the calculation by assuming that our unknown planet has the same density as Earth, \( 5500 \, \text{kg/m}^3 \).

• This assumption helps us use Earth's characteristics to estimate the unknown planet's size.
• Density affects the gravitational force a planet can exert.

Density is involved in the equation for the mass of the planet, combining volume and density:\[M = \frac{4}{3} \pi R^3 \times 5500\]where \( M \) denotes the mass of the planet, and \( R \) is its radius.However, density alone isn't enough to determine gravity or radius without additional data, like the gravitational attraction measured by the wrench's fall.

Radius Calculation

Knowing the planet's gravitational pull and assuming Earth's density, we can estimate the planet's radius. Gravitational attraction relates to mass and the planet's radius:\[g = \frac{GM}{R^2}\]We substitute the mass \( M \) using the planet's volume and density:\[M = \frac{4}{3} \pi R^3 \times 5500\]By rearranging the initial formula to solve for \( R \), we have:\[g = \frac{4}{3} \pi G \cdot R \cdot 5500\]This is a complex maneuver that merges gravitational physics with volumetric analysis.Getting \( R \) means balancing gravity's pull with planetary mass and size. Assume \( g = 15.21 \, \text{m/s}^2 \) and calculate:\[R \approx 5.18 \times 10^6 \, \text{m}\]This radius helps us imagine the planet's size, showing how measurements on gravity translate into physical dimensions.