Problem 52 - University Physics with Modern Physics

Question

A landing craft with mass \(12,500 \mathrm{kg}\) is in a circular orbit \(5.75 \times 10^{3} \mathrm{m}\) above the surface of a planet. The period of the orbit is 5800 s. The astronauts in the lander measure the diameter of the planet to be \(9.60 \times 10^{6} \mathrm{m} .\) The lander sets down at the north pole of the planet. What is the weight of an 85.6 -kg astronaut as he steps out onto the planet's surface?

Step-by-Step Solution

Answer

The weight of the astronaut is approximately 806.35 N.

Step 1: Determine the radius of the orbit

The radius of the planet is half of the diameter, so the radius \( R_p = \frac{9.60 \times 10^{6}}{2} \, \text{m} = 4.80 \times 10^{6} \, \text{m} \). The radius of the orbit \( R_o \) is the planet's radius plus the altitude of the orbit: \( R_o = 4.80 \times 10^{6} + 5.75 \times 10^{3} = 4.80575 \times 10^{6} \, \text{m} \).

Step 2: Use orbital mechanics to find gravitational force

The gravitational force acting on the lander is equal to the centripetal force required to maintain the orbit. This can be expressed as: \( F = \frac{G \, M \, m}{R_o^2} = \frac{m \, v^2}{R_o} \). Since \( v = \frac{2 \pi R_o}{T} \), substituting it back gives \( F = m \left(\frac{2\pi R_o}{T}\right)^2 / R_o = \frac{4 \pi^2 \times m \times R_o}{T^2} \).

Step 3: Solve for planet's gravitational constant

Using the formula derived in Step 2 and the known values \( m = 12500 \, \text{kg} \), \( R_o = 4.80575 \times 10^{6} \, \text{m} \), and \( T = 5800 \, \text{s} \), calculate the gravitational force: \[ F = \frac{4 \pi^2 \cdot 12500 \cdot 4.80575 \times 10^{6}}{5800^2} \approx 1.061 \times 10^8 \, \text{N} \]. Since this equals \( \frac{G \, M \, m}{R_o^2} \), solve for \( G \times M \): \[ G \times M = F \times R_o^2 / m = 1.061 \times 10^8 \times 4.80575^2 \times 10^{12} / 12500 \].

Step 4: Calculate weight of astronaut on planet

Once \( G \times M \) is found, use it to calculate the gravitational acceleration \( g \) on the surface using \( g = \frac{G \times M}{R_p^2} \). Then, the weight of the astronaut is \( W = m_a \times g \), where \( m_a = 85.6 \, \text{kg} \). Solve this: \( W = 85.6 \cdot \frac{G \times M}{(4.80 \times 10^{6})^2} \).

Step 5: Final Calculation

Plug in the values obtained from previous steps and calculate: \[ W \approx 85.6 \cdot 9.42 \, \text{m/s}^2 = 806.35 \, \text{N} \]. Calculate the final weight using the planet's gravitation.

Key Concepts

Gravitational Force

Gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another. Within the context of orbital mechanics, this force is crucial because it governs the motion of objects in space. For example, the gravitational pull that a planet exerts on an orbiting spacecraft or satellite is what keeps it in orbit.

The formula for gravitational force is given by Newton's law of universal gravitation:

• \( F = \frac{G \, M \, m}{r^2} \)

In this formula:

• \( F \) is the gravitational force.
• \( G \) is the gravitational constant, \( 6.674 \, \times\, 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
• \( M \) and \( m \) are the masses of the two objects.
• \( r \) is the distance between the centers of the two masses.

In the exercise, the gravitational force between the landing craft and the planet plays a vital role in keeping the craft in its circular orbit until it decides to land.

Centripetal Force

Centripetal force is the force required to keep a body moving in a circular path. This force acts perpendicular to the object's velocity and towards the center of the circle. In orbital mechanics, it is this force that prevents an object in orbit from flying off into space in a straight line.

For an object of mass \( m \) moving at a velocity \( v \) in a circle of radius \( r \), the centripetal force \( F_c \) required is:

• \( F_c = \frac{m \, v^2}{r} \)

In the context of planetary orbits, this centripetal force is provided entirely by the gravitational force exerted by the planet on the object. Thus, the equations for gravitational force and centripetal force are set equal in orbital problems, as was needed to solve the initial exercise.

Planetary Orbits

Planetary orbits refer to the path that planets and other celestial bodies follow due to gravitational attraction. Most orbits are elliptical, but in many cases, such as the exercise, they are approximated as circular for simplicity. The parameters defining these orbits can include:

• The orbital radius, which is the distance from the planet's center to the object.
• The orbital period, which is the time it takes for one complete trip around the planet.

Understanding orbital dynamics helps us analyze movements not just here on Earth but throughout our solar system. Factors like the planet's mass and the gravitational constant play crucial roles in determining the orbit's characteristics. The orbital mechanics examined in the exercise are foundational to space travel and satellite technology.

Gravitational Acceleration

Gravitational acceleration, often denoted \( g \), is the acceleration of an object caused by the force of gravity from a massive body like a planet. Near the surface of a planet, it determines how strong the pull of gravity is on objects. This is why two astronauts stepping on different planets will feel different strengths of gravity.

The formula to calculate gravitational acceleration at the surface of a planet is:

• \( g = \frac{G \, M}{R^2} \)

where:

• \( G \) is the gravitational constant.
• \( M \) is the mass of the planet.
• \( R \) is the radius of the planet.

Using gravitational acceleration, we can compute the weight of an object using \( W = m \, g \), where \( m \) is the object's mass. In the exercise, this calculation determined the astronaut's weight on the planet's surface after landing.