Question

The acceleration due to gravity on the Moon is \(1.62 \mathrm{~m} / \mathrm{s}^{2}\) approximately a sixth of the value on Earth. For a given initial velocity \(v_{0}\) and a given launch angle \(\theta_{0}\), the ratio of the range of an ideal projectile on the Moon to the range of the same projectile on Earth, \(R_{\text {Moon }} / R_{\text {Earth }},\) will be approximately a) 6 . b) 3. c) 12 . d) 5. e) 1 .

Short answer

Answer: The ratio of the range of a projectile on the Moon to the range of the same projectile on Earth is approximately 6.05.

Step-by-step solution

Write the Range of a Projectile formula

The formula for the range of a projectile is given by: \(R = \frac{v_{0}^{2} \cdot sin(2 \theta_{0})}{g}\), where \(v_0\) is the initial velocity, \(\theta_0\) is the launch angle, and \(g\) is the acceleration due to gravity.

Calculate the Range on Earth

Now we will substitute the acceleration due to gravity for Earth, which is \(g_{\text{Earth}} = 9.81 \,\mathrm{m/s^2}\) in the formula for the range for Earth, which is given by: \(R_{\text{Earth}} = \frac{v_{0}^{2} \cdot sin(2 \theta_{0})}{g_{\text{Earth}}}\).

Calculate the Range on the Moon

Similarly, we will substitute the acceleration due to gravity for the Moon, which is \(g_{\text{Moon}} = 1.62 \,\mathrm{m/s^2}\) in the formula for the range for the Moon, which is given by: \(R_{\text{Moon}} = \frac{v_{0}^{2} \cdot sin(2 \theta_{0})}{g_{\text{Moon}}}\).

Calculate the Range Ratio

To find the ratio of the range of the projectile on the Moon to the range of the same projectile on Earth, we'll divide the range on the Moon by the range on Earth: \(R_{\text{Moon}} / R_{\text {Earth}} = \frac{\frac{v_{0}^{2} \cdot sin(2 \theta_{0})}{g_{\text{Moon}}}}{\frac{v_{0}^{2} \cdot sin(2 \theta_{0})}{g_{\text{Earth}}}}\).

Simplify the Range Ratio expression

By simplifying the equation, we get: \(R_{\text{Moon}} / R_{\text {Earth}} = \frac{g_{\text{Earth}}}{g_{\text{Moon}}}\).

Compute the Range Ratio with acceleration due to gravity

Now we know the relationship between the projectile range ratio and acceleration due to gravity. To find the ratio, substitute the values of \(g_{\text{Earth}}\) and \(g_{\text{Moon}}\): \(R_{\text{Moon }} / R_{\text {Earth}} = \frac{9.81 \,\mathrm{m/s^2}}{1.62 \,\mathrm{m/s^2}}\).

Calculate the ratio

Divide \(9.81\) by \(1.62\) to get the approximate ratio: \(R_{\text{Moon}} / R_{\text {Earth}} \approx 6.05\). The range ratio is approximately 6.05, which is closest to option (a). So, the answer is (a) 6.

Key concepts

Acceleration Due to Gravity

Understanding the acceleration due to gravity is essential for analyzing projectile motion. It is the constant acceleration that a body experiences when in free fall, solely under the influence of gravity's pull, with no other forces acting on it. On Earth, this value is approximately 9.81 m/s², a crucial constant that allows us to predict the motion of projectiles.

However, this acceleration isn't the same everywhere—it varies with location. For example, on the Moon, the acceleration due to gravity is much less, about 1.62 m/s². The significance of this difference comes into play when comparing projectile motion across these two celestial bodies. A projectile on the Moon, affected by its lower gravity, will travel farther and stay in the air longer before landing compared to one on Earth, assuming the same initial velocity and launch angle.

Range of Projectile Motion

The range of projectile motion is defined as the horizontal distance a projectile travels before hitting the ground. This distance depends primarily on the initial velocity, the angle of launch, and importantly, the acceleration due to gravity. The formula to calculate the range is:
\[ R = \frac{v_{0}^{2} \cdot \sin(2 \theta_{0})}{g} \
where \( R \) is the range, \( v_0 \) the initial velocity, \( \theta_0 \) the launch angle, and \( g \) the acceleration due to gravity. With a smaller \( g \) value, such as on the Moon, the denominator in the range equation is smaller, leading to a larger range for the projectile.

Projectile Launch Angle

The launch angle of a projectile is the angle at which it is projected into the air relative to the horizontal. This angle, denoted by \( \theta_0 \), is a determining factor for the trajectory and range of the projectile. An angle of 45 degrees maximizes the range of the projectile in a vacuum or where air resistance is negligible. Doubling this angle results in a sin(90 degrees), which is the maximum value the sine function can attain, leading to the greatest possible range for a given initial velocity and gravitational acceleration.

Gravity on Moon vs Earth

Gravity on the Moon versus Earth affects not just the living conditions but also the physics of how objects move. The Moon's gravity is about one-sixth that of Earth's, which substantially influences projectile motion. A projectile on the Moon will have a longer duration of flight and a greater range compared to Earth for the same initial conditions.

This difference allows objects on the Moon to travel farther distances when thrown; athletes, for example, could jump higher and throw objects much further. Explorers on the Moon experience this firsthand when they traverse its dusty plains, a stark contrast to the movements limited by Earth's stronger gravitational grip.