Problem 78 - College Physics

Question

The acceleration due to gravity near a planet's surface is known to be \(3.00 \mathrm{~m} / \mathrm{s}^{2}\). If the escape speed from the planet is \(8.42 \mathrm{~km} / \mathrm{s},\) (a) determine its radius. (b) Find the mass of the planet. (c) If a probe is launched from its surface with a speed twice the escape speed and then coasts outward, neglecting other nearby astronomical bodies, what will be its speed when it is very far from the planet? (Neglect any atmospheric effects also.) (d) Under these launch conditions, at what distance will its speed be equal to the escape speed?

Step-by-Step Solution

Answer

(a) Radius \( R = 1.18 \times 10^7 \text{ m} \). (b) Mass \( M = 2.34 \times 10^{24} \text{ kg} \). (c) Speed very far \( v_f = 14580 \text{ m/s} \). (d) Distance \( d = 2.36 \times 10^7 \text{ m} \).

Step 1: Relate escape speed to gravitational potential energy

The escape speed \( v_e \) is related to the gravitational potential energy by the equation \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)), \( M \) is the mass of the planet, and \( R \) is its radius. Use this formula to express \( M \) in terms of \( R \).

Step 2: Express acceleration due to gravity in terms of mass and radius

The acceleration due to gravity \( g \) on the planet's surface is given as \( g = \frac{GM}{R^2} \). You have \( g = 3.00 \, \text{m/s}^2 \). Replace \( M \) from Step 1 into this equation to find \( R \).

Step 3: Solve for the radius \( R \) of the planet

Since \( v_e = 8.42 \, \text{km/s} = 8420 \, \text{m/s} \), substitute this value and \( g = 3.00 \, \text{m/s}^2 \) into the equation \( g = \frac{v_e^2}{2R} \). Solve for \( R \). \[ R = \frac{v_e^2}{2g} = \frac{(8420)^2}{2 \times 3.00} = 1.18 \times 10^7 \, \text{m} \]

Step 4: Calculate the mass of the planet

Now that \( R \) is known, substitute \( R = 1.18 \times 10^7 \, \text{m} \) back into the equation \( v_e = \sqrt{\frac{2GM}{R}} \) and solve for \( M \). \[ M = \frac{v_e^2 R}{2G} = \frac{(8420)^2 \times 1.18 \times 10^7}{2 \times 6.674 \times 10^{-11}} = 2.34 \times 10^{24} \, \text{kg} \]

Step 5: Determine final speed of the probe far from the planet

The probe is launched with twice the escape speed, so initial speed \( v_0 = 2 \times 8.42 \, \text{km/s} = 16.84 \, \text{km/s} = 16840 \, \text{m/s} \). Use conservation of energy: initial kinetic energy minus potential energy equals final kinetic energy since very far, potential energy is zero with negligible speed.\[ \frac{1}{2} m v_0^2 = \frac{1}{2} m v_f^2 + \, \text{Potential Energy (at escape)} \]\[ v_f = \sqrt{v_0^2 - v_e^2} = \sqrt{(16840)^2 - (8420)^2} = 14580 \, \text{m/s} \]

Step 6: Calculate the distance where speed equals escape speed

When the speed is equal to the escape speed, use the conservation of energy again: \[ \frac{1}{2} m (2v_e)^2 - \frac{GMm}{R} = \frac{1}{2} m v_e^2 - \frac{GMm}{d} \]Simplify to find \( d \): \[ d = 2R = 2 \times 1.18 \times 10^7 = 2.36 \times 10^7 \, \text{m} \]

Key Concepts

Escape Velocity

Escape velocity is the minimum speed required for an object to break free from the gravitational pull of a celestial body without further propulsion. This concept is crucial for launching spacecraft from planets.
The formula for escape velocity can be derived from the balance of kinetic energy and gravitational potential energy: \ v_{e} = \sqrt{\frac{2GM}{R}}, \ where:

• \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)),
• \( M \) is the mass of the body from which an object is trying to escape,
• \( R \) is the radius from the center of the body to the point of escape.

Understanding escape velocity helps us comprehend phenomena like why Earth’s gravitational pull is stronger compared to the Moon's, and hence requires different energies to reach space.

Planetary Mass Calculation

Determining the mass of a planet involves using both the escape velocity and the gravitational acceleration at the planet’s surface. From our given gravitational acceleration \( g = 3.00 \, \text{m/s}^2 \) and solving the equations for both escape velocity and surface gravity, we have:
Surface Gravity equation: \[ g = \frac{GM}{R^2}\]Replace \( M \) from the escape velocity formula: \[ M = \frac{v_{e}^2 R}{2G}\]By solving these together, you find the planetary mass is \( 2.34 \times 10^{24} \, \text{kg} \). This calculation is crucial for understanding the size and strength of a planet's gravitational field, which influences everything from orbital paths to atmospheric conditions.

Kinetic Energy Conservation

When a probe is launched, conservation of energy is a vital principle. Initially, the probe has significant kinetic energy as it moves with a speed twice the escape velocity. This energy gradually converts into potential energy in gravitational contexts until the probe is far enough from the planet where its speed approaches another balance point.
The kinetic energy conservation can mathematically be expressed as:\[ \frac{1}{2} m v_0^2 = \frac{1}{2} m v_f^2 + \text{Potential Energy}\]Where \( v_0 = 16840 \, \text{m/s} \) and the final speed \( v_f = 14580 \, \text{m/s} \). The conservation principle allows us to determine how energy transfers within a system remain constant, which leads to insights on probe trajectories and velocities as they leave the planet's gravitational influence.

Gravitational Potential Energy

Gravitational potential energy is the energy possessed by an object because of its position in a gravitational field, often relative to another planet or moon. It plays a crucial role in planetary sciences and space mechanics.
For an object at a distance \( d \) from a planet, the potential energy is given by \(-\frac{GMm}{d}\). This energy is negative because it takes work to move a mass \( m \) against gravity to that point. When a probe moves away from the planet, this potential energy changes.
By working through these principles and using equations such as:\[ \text{Energy at } R = \text{Energy at } d\]we deduce the distance \( d \) at which speed equals the initial escape speed. This technical insight is important for missions determining escape paths and safe distances to release payloads or reposition satellites.