Problem 93 - College Algebra and Calculus: An Applied Approach

Question

The escape velocity (in meters per second) on the moon is \(\sqrt{\frac{2\left(6.67 \times 10^{-11}\right)\left(7.36 \times 10^{22}\right)}{1.74 \times 10^{6}}}\) If all the fuel is consumed during launching, will a rocket with an initial velocity of 2000 meters per second escape the gravitational field of the moon?

Step-by-Step Solution

Answer

No, the rocket will not escape the moon's gravitational field. Its initial velocity of 2000 m/s is less than the moon's escape velocity of approximately 2380 m/s.

Step 1: Calculate the escape velocity from the moon's surface

To calculate the escape velocity, we plug into the given formula the values for the gravitational constant \(6.67 \times 10^{-11}\) N(m/kg)^2, the mass of the moon \(7.36 \times 10^{22}\) kg, and the radius of the moon \(1.74 \times 10^{6}\) m. We then implement the square root operation. The escape velocity \(v_e\) is thus found to be \(v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (7.36 \times 10^{22})}{1.74 \times 10^{6}}}\)

Step 2: Evaluate the expression for escape velocity

We need to carefully perform the multiplication and division operations, as well as the square root operation. When calculating this expression, we find that the escape velocity \(v_e\) is approximately \(2380\) m/s.

Step 3: Compare the escape velocity with the rocket's initial velocity

In this step, we compare the rocket's initial velocity of 2000 m/s with the escape velocity. As the initial velocity of the rocket is less than the escape velocity from the moon's surface, it will not be able to escape the moon's gravitational field if all the fuel is consumed during launching.

Key Concepts

Gravitational Field

A gravitational field exists around any mass, like a planet or moon, and this field pulls objects towards it. Every mass generates a gravitational pull which is proportional to the mass itself. On the moon, this field is weaker than Earth's due to its smaller mass.
As you move away from the surface of the moon, the gravitational pull weakens. This is why objects, like rockets, need to reach a certain speed to break free from this pull. This necessary speed is known as escape velocity. The gravitational field ensures that objects need sufficient energy to escape into space.

• The intensity of the gravitational field is directly related to the mass of the body.
• The escape velocity is higher on Earth than on the moon, due to its stronger gravitational field.
• Rockets need enough initial velocity to overcome the gravitational pull of the moon.

Initial Velocity

Initial velocity is the speed that an object, like a rocket, begins its journey with. For a rocket attempting to escape a celestial body's gravity, this is the starting speed right from the launch.
In our exercise, the rocket's initial velocity is 2000 meters per second. This is the speed that the rocket begins with before trying to escape the moon’s gravitational field.

• If this initial speed is less than the required escape velocity, the object will fall back.
• An object's initial velocity plays a crucial role in determining if it can escape the gravitational pull.
• A higher initial velocity than the escape velocity means the rocket can leave the moon's gravitational influence.

Gravitational Constant

The gravitational constant, often denoted by the symbol \(G\), is a key part of Newton’s law of universal gravitation. It tells us how strong the gravitational force is between two masses.
In physics, \(G\) has a value of approximately \(6.67 \times 10^{-11}\) N(m/kg)^2. This constant is essential when calculating the force of gravity between celestial bodies, like the moon and an object trying to leave its surface.

• \(G\) is used to calculate gravitational forces in space and on Earth.
• Without the gravitational constant, we couldn't calculate the escape velocity accurately.
• \(G\) is important for understanding how planets, stars, and other bodies interact through gravity.