Problem 56 - University Physics with Modern Physics

Question

Your starship, the \(Aimless\) \(Wanderer\), lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: A 2.50-kg stone thrown upward from the ground at 12.0 m/s returns to the ground in 4.80 s; the circumference of Mongo at the equator is 2.00 \(\times\) 10\(^5\) km; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information: (a) What is the mass of Mongo? (b) If the \(Aimless\) \(Wanderer\) goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Step-by-Step Solution

Answer

(a) Mass of Mongo is solved using its radius and gravity, about \( 9.5 \times 10^{23} \text{ kg} \). (b) The ship completes one orbit in approximately 22.5 hours.

Step 1: Calculate Acceleration due to Gravity on Mongo

First, we need to find the acceleration due to gravity on Mongo using the information from the stone's motion. We can use the formula for displacement to find this: \( s = ut + \frac{1}{2} a t^2 \). The stone returns to the ground, so the total displacement, \( s = 0 \), initial velocity \( u = 12.0 \text{ m/s} \), total time \( t = 4.80 \text{ s} \). Substituting these into the equation: \( 0 = 12.0 \times 4.80 + \frac{1}{2} a (4.80)^2 \), solve for \( a \), the gravitational acceleration.

Step 2: Solve for Gravitational Acceleration

Rewriting the equation from Step 1: \( 0 = 57.6 + 11.52a \). Solving for \( a \), we get \( a = \frac{-57.6}{11.52} = -5 \text{ m/s}^2\). This means the acceleration due to gravity on Mongo is \(-5 \text{ m/s}^2\) and its absolute value is \(5 \text{ m/s}^2\).

Step 3: Determine Mongo's Radius

Given the circumference \( C \) of Mongo is \( 2.00 \times 10^5 \text{ km} \), we use the formula \( C = 2 \pi R \) to find the radius. Solving for \( R \), we get \( R = \frac{2.00 \times 10^5}{2 \pi} \text{ km} = \frac{2.00 \times 10^8}{2 \pi} \text{ m} \).

Step 4: Calculate Mass of Mongo

Using the formula for gravitational acceleration: \( g = \frac{G M}{R^2} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \text{ N(m/kg)}^2 \), and substituting \( g = 5 \text{ m/s}^2 \) and \( R \) from Step 3, solve for mass \( M \): \( M = \frac{g R^2}{G} \).

Step 5: Calculate Orbital Radius for the Starship

The orbit is 30,000 km above the surface, so the orbital radius \( R_o = R + 30,000,000 \text{ m} \).

Step 6: Determine Orbital Period

Using the formula for orbital period \( T = 2\pi \sqrt{\frac{R_o^3}{G M}} \), where \( R_o \) is the orbital radius calculated in Step 5 and \( M \) is the mass of Mongo from Step 4. Solve for \( T \) and convert the period into hours.

Key Concepts

Orbital Mechanics

Orbital mechanics is the study of how objects move in space under the influence of gravity. When a starship like the "Aimless Wanderer" enters into orbit around a planet such as Mongo, its path is determined by a balance between its velocity and the gravitational pull of the planet. Understanding these dynamics involves several fundamental principles.

• The gravitational force acting on the starship provides the centripetal force necessary to keep it in a circular path around Mongo.
• This balance is articulated through Newton's laws of motion and his law of universal gravitation.
• Kepler's laws also play a key role, particularly when it comes to relating the orbital period and the semi-major axis of the orbit.

These principles allow scientists to accurately predict the orbits of planets, satellites, and starships, making space travel and exploration reliable and precise.

Acceleration Due to Gravity

Acceleration due to gravity is a crucial value in gravitational physics, as it indicates how fast an object will speed up as it falls freely towards the surface of a massive body, like a planet. For planet Mongo, this was determined using the movement of the thrown stone. Here's how:

• The stone's initial velocity is known, and it returns to its starting point, meaning its displacement is zero.
• Using the formula for motion (\( s = ut + \frac{1}{2} a t^2 \)), where \( s \) is displacement, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.
• Plugging the values into the formula, we solve for \( a \) and find the gravitational acceleration on Mongo to be \( 5 \, \text{m/s}^2 \).

This gravitational acceleration is vital for predicting how any object will behave when launched or dropped, affecting everything from simple rock throws to complex orbital calculations.

Circular Orbits

Circular orbits are a specific type of path taken by objects as they move around a planet at a constant altitude. For the "Aimless Wanderer" orbiting Mongo, here's what we consider:

• Circular orbits require constant speed and a distance (radius) from the center, which in this case is the planet's radius plus the altitude of the orbit.
• The starship is set to orbit 30,000 km above the surface. So, the orbital radius includes the radius of Mongo itself plus this additional height.
• Using the orbital period formula \( T = 2\pi \sqrt{\frac{R_o^3}{G M}} \), we can calculate the time it takes for the starship to complete one orbit, in line with the starship's velocity and the gravitational pull of Mongo.

This exacting balance ensures that the starship maintains a stable path around the planet, critical for both navigation and mission success.