Problem 43 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Suppose the space shuttle is in orbit \(400 \mathrm{~km}\) from the Earth's surface, and circles the Earth about once every 90 min. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of \(g,\) the gravitational acceleration at the Earth's surface.

Step-by-Step Solution

Answer

The centripetal acceleration is approximately 0.92g.

Step 1: Analyze the Problem Requirements

We need to determine the centripetal acceleration of a space shuttle orbiting the Earth 400 km above the surface. We'll express this acceleration in terms of the gravitational acceleration \( g \) at the Earth's surface.

Step 2: Determine Shuttle's Orbital Radius

The Earth's radius is approximately \( 6371 \text{ km} \). The orbit is 400 km above the surface, so the orbital radius \( R \) is \( 6371 \text{ km} + 400 \text{ km} = 6771 \text{ km} \) or \( 6771 \times 10^3 \text{ m} \).

Step 3: Calculate Orbital Velocity

The shuttle completes an orbit in 90 minutes (or \( 5400 \text{ seconds} \)). The formula for velocity \( v \) in a circular orbit is \( v = \frac{2\pi R}{T} \), where \( R \) is the orbital radius and \( T \) is the orbital period. Substitute \( R = 6771 \times 10^3 \text{ m} \) and \( T = 5400 \text{ s} \).

Step 4: Compute Centripetal Acceleration

Centripetal acceleration \( a_c \) is given by \( a_c = \frac{v^2}{R} \). Use the velocity from the previous step and orbit radius to calculate \( a_c \).

Step 5: Express in Terms of Gravitational Acceleration \( g \)

To express the centripetal acceleration \( a_c \) in terms of \( g \), use \( g = 9.8 \text{ m/s}^2 \). The calculated \( a_c \) will be compared to \( g \), and the final answer will be in terms of \( g \).

Key Concepts

Orbital Radius

To understand how the space shuttle maintains its orbit, we must first grasp the concept of **orbital radius**. This defines the distance from the center of the Earth to the space shuttle in its orbit.
For the space shuttle, this orbital radius isn't just the height above the Earth's surface. It includes the Earth's radius as well. The Earth has a radius of about 6371 km.
If the space shuttle is orbiting 400 km above the surface:

• Orbital Radius (\( R \)) = Earth's Radius + Height above Earth
• \( R = 6371 ext{ km} + 400 ext{ km} = 6771 ext{ km} \)
• To use in calculations, convert this to meters: \( 6771 imes 10^3 ext{ m} \)

The orbital radius plays a critical role in determining both the shuttle's velocity and the centripetal acceleration required to maintain its orbit.

Gravitational Acceleration

Gravitational acceleration (\( g \)) is a measure of the force of gravity acting on objects at Earth's surface. On average, it is approximately 9.8 m/s², but this force diminishes with altitude.
While the space shuttle is in orbit, it is much farther from Earth and thus experiences reduced gravitational forces compared to what we feel on the surface.

• Gravitational force decreases with distance from the Earth's center.
• Despite this reduction, gravity still acts as a key force, keeping the shuttle in its curved orbit.
• This is necessary for maintaining the centripetal acceleration that keeps the shuttle following its orbital path instead of flying off into space.

When calculating the shuttle's centripetal acceleration, we can express the result in terms of \( g \), effectively comparing how strong the necessary acceleration is relative to surface gravity.

Orbital Velocity

The shuttle's **orbital velocity** is the speed it must maintain to stay in a stable orbit, balancing between gravitational pull and the inertia of its motion.
The formula to calculate orbital velocity \( (v) \) is:

• \( v = \frac{2\pi R}{T} \)
• \( R \) is the orbital radius (6771 km or \( 6771 \times 10^3 \text{ m} \)
• \( T \) is the orbital period (90 minutes or 5400 seconds).

By substituting these values, the velocity ensures the shuttle circles the Earth without veering off its path:

• Maintaining the correct orbital velocity counteracts gravity enough to sustain orbit.
• If the shuttle’s speed were too high or too low, it would either ascend away from the Earth or fall towards it, respectively.

This calculation is crucial as it informs how fast the shuttle must travel to achieve the balance required for a stable orbit.