Problem 113 - Algebra A Combined Function

Question

Escape velocity is the minimum speed that an object must reach to escape a planet's pull of gravity. Escape velocity \(v\) is given by the equation \(v=\sqrt{\frac{2 G m}{r}},\) where \(m\) is the mass of the planet, \(r\) is its radius, and \(G\) is the universal gravitational constant, which has a value of \(G=6.67 \times 10^{-11} \mathrm{~m}^{3} / \mathrm{kg} \cdot \mathrm{s}^{2} .\) The mass of Earth is \(5.97 \times 10^{24} \mathrm{~kg},\) and its radius is \(6.37 \times 10^{6} \mathrm{~m}\). Use this information to find the escape velocity for Earth in meters per second. Round to the nearest whole number. (Source: National Space Science Data Center)

Step-by-Step Solution

Answer

Earth's escape velocity is approximately 11186 m/s.

Step 1: Identify the given values

We know from the problem that the universal gravitational constant \(G = 6.67 \times 10^{-11} \, \mathrm{m}^3/\mathrm{kg}\cdot\mathrm{s}^2\). The mass of Earth \(m = 5.97 \times 10^{24} \, \mathrm{kg}\), and the radius of Earth \(r = 6.37 \times 10^6 \, \mathrm{m}\).

Step 2: Write down the escape velocity formula

The formula for escape velocity \(v\) is given by \(v = \sqrt{\frac{2Gm}{r}}\).

Step 3: Substitute the given values into the formula

Replace \(G\), \(m\), and \(r\) in the formula with the given values: \[ v = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \mathrm{m}^3/\mathrm{kg}\cdot\mathrm{s}^2 \times 5.97 \times 10^{24} \, \mathrm{kg}}{6.37 \times 10^6 \, \mathrm{m}}} \]

Step 4: Simplify the expression under the square root

First, calculate the product of \(2 \cdot 6.67 \times 10^{-11}\) and \(5.97 \times 10^{24}\). This product is: \[ 7.9746 \times 10^{14} \, \mathrm{m}^3/\mathrm{s}^2 \]

Step 5: Divide by the radius

Now, divide \(7.9746 \times 10^{14} \, \mathrm{m}^3/\mathrm{s}^2\) by \(6.37 \times 10^6 \, \mathrm{m}\) to get \[ \frac{7.9746 \times 10^{14}}{6.37 \times 10^6} = 1.25227 \times 10^8 \, \mathrm{m}^2/\mathrm{s}^2 \]

Step 6: Calculate the square root

Find the square root of the value found in the previous step: \[ v = \sqrt{1.25227 \times 10^8} \approx 11186 \]

Step 7: Round to the nearest whole number

Round the calculated escape velocity to the nearest whole number. Thus, the escape velocity \(v\) for Earth is approximately 11186 m/s.

Key Concepts

Universal Gravitational Constant

The Universal Gravitational Constant, denoted as \( G \), is a critical factor in the law of universal gravitation. It quantifies the strength of gravity across the universe, acting as a proportionality constant in the gravitational force equation. This constant is expressed in units of \( \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{s}^2 \), highlighting its role in gravitational calculations. The value of \( G \) stands at \( 6.67 \times 10^{-11} \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{s}^2 \), a value derived from precise experimental measurements. It demonstrates the relatively weak nature of gravitational force compared to other fundamental forces.

• This constant is essential for calculating gravitational forces between two masses.
• It is a universal constant, meaning it applies to all objects everywhere in the universe.

The presence of \( G \) in the escape velocity formula emphasizes how gravity limits an object's ability to escape a planet's gravitational pull.

Mass of Earth

The mass of Earth is a fundamental detail impacting the gravitational force exerted by the planet. It is approximately \( 5.97 \times 10^{24} \, \mathrm{kg} \). This colossal mass generates the gravitational pull that keeps objects, including satellites and the atmosphere, bound to the Earth. Understanding Earth's mass is crucial for several reasons:

• It determines the strength of Earth's gravity.
• It plays a direct role in calculating the escape velocity.
• The mass influences Earth's orbital characteristics.

The mass, combined with the radius of Earth, enables us to determine the escape velocity necessary for an object to break free from Earth's gravitational field.

Radius of Earth

The radius of Earth is another critical parameter in gravitational calculations, especially concerning escape velocity. Measured from the center of Earth to its surface, the radius is approximately \( 6.37 \times 10^6 \mathrm{~m} \). This measurement is vital because:

• It directly affects the gravitational pull experienced by objects on Earth's surface.
• The radius is used in equations that relate to the gravitational force and escape velocity.
• A larger radius would mean a smaller gravitational force at the surface, affecting escape velocity.

In the context of escape velocity, the radius works in conjunction with the mass and the universal gravitational constant to define how fast an object must travel to escape Earth's atmosphere.

Calculating Escape Velocity

Escape velocity is the speed needed for an object to break free from a planet's gravitational field without further propulsion. It signifies the minimum launch speed an object must reach to overcome gravitational pull.The escape velocity \( v \) formula is given by:\[v = \sqrt{\frac{2 G m}{r}}\]in which:

• \( G \) is the universal gravitational constant \( 6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{kg} \cdot \mathrm{s}^2 \)
• \( m \) is the mass of the planet (Earth: \( 5.97 \times 10^{24} \mathrm{~kg} \))
• \( r \) is the radius of the planet (Earth: \( 6.37 \times 10^6 \mathrm{~m} \))

Applying these values, you can determine that Earth's escape velocity is approximately 11186 meters per second. This value represents the speed required to leave Earth's gravitational influence without additional energy input.