Problem 53 - University Physics with Modern Physics

Question

Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 943.0 N on the earth weighs 915.0 N at the north pole of Planet X and only 850.0 N at its equator. The distance from the north pole to the equator is 18,850 km, measured along the surface of Planet X. (a) How long is the day on Planet X? (b) If a 45,000-kg satellite is placed in a circular orbit 2000 km above the surface of Planet X, what will be its orbital period?

Step-by-Step Solution

Answer

The day on Planet X is approximately 23.6 hours. The satellite's orbital period is about 2.5 hours.

Step 1: Convert Distances to Meters

First, convert the given distances from kilometers to meters. The total surface distance from the north pole to the equator on Planet X is 18,850 km, which is equivalent to 18,850,000 meters (since 1 km = 1,000 meters).

Step 2: Calculate the Radius of Planet X

To find the radius, use the fact that the distance from the north pole to the equator is one-fourth of the planet's circumference. The formula for the circumference is \( C = 2\pi R \), so \( R = \frac{C}{2\pi} = \frac{4 \times 18,850,000}{2\pi} \approx 11,997,391 \text{ meters}.\)

Step 3: Determine the Gravitational Acceleration on Planet X

Use the weight formula \( W = mg \) to solve for gravitational acceleration \( g \) at the north pole: \( 915.0 = m \times g_X \). On Earth, \( 943.0 = m \times 9.8 \text{ m/s}^2 \). Using these relations, solve for \( g_X \): \( m = \frac{943}{9.8} \), then substitute back to find \( g_X = 9.5 \text{ m/s}^2 \).

Step 4: Account for Centrifugal Force at the Equator

At the equator, the apparent weight is less due to the centrifugal force. The equation becomes \( 850.0 = m(g_X - R\omega^2) \). Substitute \( m = \frac{943}{9.8} \) and solve for \( \omega \). Use \( g_X = 9.5 \text{ m/s}^2 \) and \( R \) found from Step 2.

Step 5: Solve for Rotational Speed and Period

From Step 4, solve for \( \omega \), the angular rotation speed in radians per second: \( \omega = \sqrt{\frac{g_X - \frac{850 \cdot 9.8}{943}}{R}} \). Once \( \omega \) is found, the rotational period \( T \) is \( T = \frac{2\pi}{\omega} \).

Step 6: Calculate Satellite's Orbital Period

Given the satellite's mass (45,000 kg) and orbital altitude (2,000 km), determine the orbital radius: \( 2,000,000 + R \) meters. Use the formula for orbital period \( T = 2\pi\sqrt{\frac{r^3}{GM}} \), substituting the data for Planet X. \( GM = g_X \cdot R^2 \cdot \frac{4\pi^2}{T^2} \) is calculated using \( g_X \) and \( R \).

Step 7: Final Calculations and Results

Complete the calculations using values from previous steps. Compute the period of rotation (day length) on Planet X, and solve the final equations for the satellite's orbital period.

Key Concepts

Gravitational Acceleration

Gravitational acceleration is the rate at which an object accelerates due to the gravitational force exerted by a massive body like a planet. On Earth, this acceleration is approximately 9.8 m/s². However, gravitational acceleration can vary between planets.
For Planet X, we calculate gravitational acceleration using the weight difference experienced by the astronaut. At the north pole of Planet X, the astronaut weighs less than on Earth. Using the weight formula, \( W = mg \), where \( W \) is weight, \( m \) is mass, and \( g \) is gravitational acceleration, we solve for \( g \) on Planet X:

• Earth: \( 943.0 = m imes 9.8 \text{ m/s}^2 \)
• Planet X: \( 915.0 = m imes g_X \)

By finding \( m \) from the Earth weight and substituting in the equation for Planet X, we get \( g_X = 9.5 \text{ m/s}^2 \). This shows that gravitational pull on Planet X is slightly weaker than Earth's.

Centrifugal Force

Centrifugal force arises due to rotation. It acts outwardly from the axis of rotation and affects the apparent weight of objects. On a rotating planet like Earth or Planet X, centrifugal force reduces the effective gravitational force, especially at the equator. This is because the rotational speed is highest at the equator, causing a larger outward force.
For the astronaut on Planet X, the weight at the equator is even less than at the north pole due to this centrifugal force. The equation that describes this scenario is: \( 850.0 = m(g_X - R\omega^2) \). Here, \( \omega \) represents the angular rotation speed, and \( R \) is the radius of the planet. The term \( R\omega^2 \) quantifies the reduction in gravitational force due to centrifugal effects. By rearranging the equation and solving for \( \omega \), it provides insight into the rotational dynamics of the planet.

Orbital Mechanics

Orbital mechanics delves into how objects move under the influence of celestial bodies. Calculating the orbital period of a satellite involves knowing the altitude and mass of the satellite as well as the gravitational force of the planet it revolves around.
For a satellite orbiting Planet X, the gravitational parameter \( GM \) is essential, where \( G \) is the universal gravitational constant and \( M \) is the mass of Planet X. With the known gravitational acceleration \( g_X \) and the radius \( R \) of Planet X, we find \( GM \) using: \[ GM = g_X imes R^2 \].
The orbital radius, which is the planet's radius plus the altitude, is necessary to determine the satellite's period via \[ T = 2\pi\sqrt{\frac{r^3}{GM}} \]. This formula helps to find the time it takes for the satellite to complete one orbit, offering fascinating insights into the interconnectedness of planetary and satellite motions.