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, $R_{\text{Moon }}/R_{\text{Farth }}$ will be a) $6\mathrm{m}$ d) $5\mathrm{m}$ b) $3\mathrm{m}$ e) $1\mathrm{m}$ c) $12\mathrm{m}$

Short answer

Answer: The ratio is approximately equal to 6.

Step-by-step solution

Find the time of flight on Earth and the Moon

First we need to find the time of flight on both the Moon and Earth. The time of flight is the time it takes for a projectile to reach the ground after being launched. For both cases, we can use the following equation: $$t=\frac{2v_0\sin {\theta }_0}{g}$$ Here, $t$ is the time of flight, $v_0$ is the initial velocity, ${\theta }_0$ is the launch angle, and $g$ is the acceleration due to gravity. The time of flight on Earth, $t_E$, can be found using the Earth's acceleration due to gravity, $g_E=9.81\mathrm{m}/{\mathrm{s}}^2$: $$t_E=\frac{2v_0\sin {\theta }_0}{g_E}$$ The time of flight on the Moon, $t_M$, can be found using the Moon's acceleration due to gravity, $g_M=1.62\mathrm{m}/{\mathrm{s}}^2$: $$t_M=\frac{2v_0\sin {\theta }_0}{g_M}$$

Find the range on Earth and the Moon

Next, we can find the range of the projectile on both Earth and the Moon. The range can be calculated using the following equation: $$R=v_0\cos {\theta }_0\cdot t$$ Here, $R$ is the range, $v_0$ is the initial velocity, ${\theta }_0$ is the launch angle, and $t$ is the time of flight. The range on Earth, $R_E$, can be calculated using the time of flight on Earth, $t_E$: $$R_E=v_0\cos {\theta }_0\cdot t_E$$ Similarly, the range on the Moon, $R_M$, can be calculated using the time of flight on the Moon, $t_M$: $$R_M=v_0\cos {\theta }_0\cdot t_M$$

Calculate the ratio of ranges

Now we can calculate the ratio of the range of an ideal projectile on the Moon to the range of the same projectile on Earth: $$\frac{R_{\text{Moon}}}{R_{\text{Earth}}}=\frac{R_M}{R_E}=\frac{v_0\cos {\theta }_0\cdot t_M}{v_0\cos {\theta }_0\cdot t_E}=\frac{t_M}{t_E}$$ We can substitute the expressions for $t_M$ and $t_E$ we found in Step 1: $$\frac{R_{\text{Moon}}}{R_{\text{Earth}}}=\frac{\frac{2v_0\sin {\theta }_0}{g_M}}{\frac{2v_0\sin {\theta }_0}{g_E}}$$ The $2v_0\sin {\theta }_0$ terms cancel out: $$\frac{R_{\text{Moon}}}{R_{\text{Earth}}}=\frac{g_E}{g_M}$$

Find the numerical value of the ratio

Finally, we can substitute the values for $g_E$ and $g_M$ to find the numerical value of the ratio: $$\frac{R_{\text{Moon}}}{R_{\text{Earth}}}=\frac{9.81\mathrm{m}/{\mathrm{s}}^2}{1.62\mathrm{m}/{\mathrm{s}}^2}$$ The ratio is approximately equal to 6. Therefore, the correct answer is a) $6\mathrm{m}$.

Key concepts

Acceleration Due to Gravity

Understanding the acceleration due to gravity is crucial when studying projectile motion, as it is one of the primary forces affecting a projectile's trajectory. On Earth, the acceleration due to gravity, denoted as $g_E$, is approximately $9.81{\text{m/s}}^2$. However, the Moon's gravitational pull is weaker, leading to an acceleration due to gravity, $g_M$, of about $1.62{\text{m/s}}^2$, which is roughly a sixth of Earth’s. This significant difference plays a pivotal role in how projectiles move on the Moon versus on Earth. For example, if you were to jump on the Moon, you would be able to leap much higher and stay in the air longer due to the lower acceleration due to gravity.

When comparing projectile motion on the two celestial bodies, the lower gravity on the Moon allows a projectile to have a greater time of flight and range for the same initial velocity and launch angle. This concept can be visualized by imagining throwing a ball; on the Moon, the ball would travel further before hitting the ground due to the weaker pull of gravity.

Time of Flight

The time of flight is the total duration a projectile remains airborne. It depends on the initial velocity, launch angle, and acceleration due to gravity. The formula for time of flight is $t=\frac{2v_0\sin {\theta }_0}{g}$, where $t$ represents the time of flight, $v_0$ the initial velocity, ${\theta }_0$ the launch angle, and $g$ the acceleration due to gravity. On the Moon, with its lower $g_M$, the time of flight increases when compared to Earth.

This implies that a basketball thrown with the same force will stay in the air longer on the Moon than it would on Earth. Such an extended time of flight can impact various aspects of projectile motion, like the range or the maximum height reached by the projectile. Students must remember that the time of flight is independent of the horizontal component of the initial velocity; it is influenced only by the vertical component and gravity.

Projectile Range

The range of a projectile is the horizontal distance it covers from its point of release to when it lands. This distance is given by the formula $R=v_0\cos {\theta }_0\cdot t$, where $R$ is the range, $v_0$ the initial velocity, ${\theta }_0$ the launch angle, and $t$ the time of flight. Since the time of flight on the Moon is extended due to the lower gravity, the range of a projectile for the same initial velocity and launch angle is also greater on the Moon than on Earth.

It's interesting to note that the range of a projectile is maximized at a launch angle of 45 degrees, assuming no air resistance. This is true both on Earth and the Moon. However, due to the lower gravity on the Moon, the same projectile will travel six times further. This is key information for professionals ranging from athletes understanding the dynamics of their motions to engineers designing lunar rovers.

Launch Angle

The launch angle, ${\theta }_0$, of a projectile is the angle at which it is released with respect to the horizontal. This angle significantly influences the shape and length of the projectile's trajectory. The optimal launch angle for the maximum range without air resistance is typically 45 degrees; however, the exact angle required to hit a specified target will vary based on different factors such as initial speed and height of release. Interestingly, while the optimal angles remain the same, the outcome in terms of range and height significantly differ between Earth and the Moon due to the different gravitational accelerations.

For instance, a lunar astronaut playing golf would find that the ball flies along a much flatter trajectory than it would on Earth for lower launch angles due to the moon's weaker gravity. It is this angle, coupled with initial velocity and local gravity, that determines the ultimate path taken by a projectile, whether it be a golf ball on the Moon or a satellite launched into Earth's orbit.