Problem 56 - University Physics with Modern Physics

Question

Planet \(\mathrm{X}\) 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 \(\mathrm{N}\) on the earth weighs 915.0 \(\mathrm{N}\) at the north pole of Planet \(\mathrm{X}\) and only 850.0 \(\mathrm{N}\) at its equator. The distance from the north pole to the equator is \(18,850 \mathrm{km},\) measured along the surface of Planet \(\mathrm{X}\) . (a) How long is the day on Planet \(\mathrm{X}\) ? (b) If a 45,000-kg satellite is placed in a circular orbit 2000 \(\mathrm{km}\) above the surface of Planet \(\mathrm{X},\) what will be its orbital period?

Step-by-Step Solution

Answer

Planet X's day is about 13.7 hours long, and the satellite's orbital period is about 4.17 hours.

Step 1: Identify the forces at the poles and equator

First, we identify the forces acting on an astronaut at the poles and the equator of Planet X. At the north pole, there is no centripetal force acting, hence the astronaut's weight of 915 N is purely gravitational force. At the equator, the astronaut's weight of 850 N is the result of gravitational force minus the centripetal force due to the planet's rotation.

Step 2: Determine gravitational acceleration

Given that the astronaut weighs 915 N at the north pole and 943 N on Earth, the gravitational acceleration on Planet X, denoted as \( g_X \), can be calculated using the formula \( g_X = \frac{W}{m} \), where \( W = 915 \) N and \( m = \frac{943}{9.81} \) kg. Solving gives \( g_X = \frac{915}{\frac{943}{9.81}} \approx 9.517 \, \text{m/s}^2 \).

Step 3: Determine centripetal acceleration at the equator

At the equator, the 850 N weight is the net force after subtracting the centripetal force, meaning \( 850 = mg_X - F_{c} \). The centripetal force \( F_c = m\omega^2 R \), where \( R \) is the radius and \( \omega = \frac{2\pi}{T} \) is the angular speed. Rearranging gives \( F_{c} = mg_X - 850 \approx 93 \, \text{N} \).

Step 4: Calculate the radius of Planet X

The radius \( R \) of Planet X is determined by dividing the distance from pole to equator (18,850 km) by \( \pi/2 \). Converting km to meters gives \( R = \frac{18,850,000}{\pi/2} \approx 12,000,200.85 \, \text{m} \).

Step 5: Calculate length of the day on Planet X

Use the centripetal force formula from Step 3 and solve for the period \( T \). With \( F_c = m \omega^2 R = 93 \) N and \( \omega = \frac{2\pi}{T} \), solve for \( T \):\[ T = 2\pi \sqrt{\frac{mR}{F_c}}. \]Plug in \( R \approx 12,000,200.85 \, \text{m} \), \( F_c = 93 \) N, and \( m = \frac{850}{9.517} \) kg to get \( T \approx 49,367 \, \text{s} \approx 13.7 \, \text{hours}. \)

Step 6: Calculate gravitational force on the satellite

For the satellite orbiting at 2000 km above the surface, its total distance from the center of Planet X is \( 12,000,200.85 + 2,000,000 \approx 14,000,200.85 \, \text{m} \). The gravitational force \( F_{g} \) it experiences is given by \( F_{g} = \frac{G M m}{R_{total}^2} = m g_{surface} \left(\frac{R_{surface}}{R_{total}}\right)^2 \).

Step 7: Determine satellite's orbital velocity and period

The orbital velocity \( v \) is determined by the balance of gravitational and centripetal forces: \( F_g = \frac{mv^2}{R_{total}} \). Solving for \( v \) and using \( v = \frac{2\pi R_{total}}{T_{orbit}} \), we find:\[ T_{orbit} \approx 2\pi \sqrt{\frac{R_{total}^3}{G M}} \approx 15,000 \, \, \text{seconds} \approx 4.17 \, \text{hours}. \]

Key Concepts

Centripetal Force

When a planet rotates, it causes objects on its surface to experience a force directed towards the center of the planet. This force is known as centripetal force. At the equator of a rotating planet like Planet X, this force is significant. It results from the planet's rotation and tends to "throw" objects outward, opposing the gravitational pull that draws objects inward.
Centripetal force can be calculated using the formula:

• \( F_c = m \omega^2 R \)

where \( m \) is the mass of the object, \( \omega \) is the angular velocity, and \( R \) is the radius of the planet. At the equator, this force decreases the effective weight of objects. This is why the astronaut weighs less at the equator of Planet X compared to the North Pole.

Gravitational Acceleration

Gravitational acceleration is the acceleration due to gravity that a planet exerts on objects at its surface. On Planet X, gravitational acceleration can be determined by the weight of an object at the poles where centripetal force has no effect.
The gravitational acceleration \( g_X \) can be calculated from the formula:

• \( g_X = \frac{W}{m} \)

Here, \( W \) is the weight of the object, and \( m \) is the mass calculated based on the weight on Earth. This value is critical for understanding how much force is necessary to keep an object stationary at the surface of a planet.

Orbital Period

The orbital period is the time a satellite takes to complete one full orbit around a planet. This depends on the radius of the orbit and the gravitational force acting on the satellite.
For a satellite orbiting 2000 km above the surface of Planet X, its orbital period \( T_{orbit} \) can be found using the formula:

• \( T_{orbit} = 2\pi \sqrt{\frac{R_{total}^3}{G M}} \)

where \( R_{total} \) is the total orbital radius, \( G \) is the gravitational constant, and \( M \) is the mass of Planet X. This formula helps determine the necessary speed and trajectory for stable satellite orbits.

Satellite Orbit

A satellite orbiting a planet moves in a path defined by the gravitational pull of the planet. In a circular orbit, the satellite's distance from the planet remains constant. On Planet X, a satellite orbiting at 2000 km above the surface will follow a path influenced by both gravity and the centripetal force required to keep it in orbit.
The satellite's velocity \( v \) can be balanced by:

• \( F_g = \frac{mv^2}{R_{total}} \)

This condition ensures that the satellite remains in a stable orbit, moving consistently around the planet while being subjected to minimal external disturbances.

Weight Variation

Weight variation refers to the change in the gravitational force experienced by an object at different locations on a planet. On Planet X, the astronaut's weight differs between the poles and the equator.
This is due to the impact of centripetal force at the equator, which decreases the effect of gravitational pull. The difference in weight is calculated using:

• Effective weight at equator = Weight at poles - Centripetal force

Such variations are essential for understanding environmental and living conditions on different parts of a rotating planet.

Spherical Geometry

Understanding the shape and dimensions of a planet is essential, especially when calculating distances and paths, such as in orbital mechanics. Planet X is described as perfectly spherical, meaning its radius can be consistently calculated from any point.
The radius \( R \) of Planet X is found using the distance from the pole to the equator divided by half of \( \pi \). This spherical shape simplifies calculations related to rotation, gravity, and orbits. Recognizing the significance of spherical geometry is important in physics and planetary studies, as it helps provide accurate measurements for various celestial mechanics.