# # Physics Behind Asteroid M's Fate in X-Men 97

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## Chapter 1: Understanding Asteroid M

Everyone is currently tuning into X-Men 97 on Disney Plus, and if you haven’t caught the season finale yet, here’s your SPOILER ALERT! Don’t worry; I’m primarily focusing on the physics involved. I enjoy merging topics I love with scientific insights—it's an enjoyable endeavor.

For those unfamiliar with the episode, Asteroid M serves as Magneto’s space base, described as “several cubic miles in size.” It doesn’t seem to be in low Earth orbit; rather, it appears to be hovering, likely sustained by some energy source (though I could be mistaken). In this episode, missiles are launched at the base, causing it to begin its descent toward Earth. That’s the essential backdrop before diving into the physics.

### Terminal Velocity

One notable quote from the episode states: “The asteroid is reaching terminal velocity.” But what does terminal velocity mean, and is it feasible for Asteroid M to achieve this speed? Let's start by considering a simple scenario. Imagine dropping a spherical rock with a radius of 1 centimeter from a high altitude. As it descends, two forces come into play:

- The gravitational force from the Earth.
- The air resistance opposing its motion.

The gravitational force is constantly pulling down, dependent on the object’s mass (m) and the gravitational field (g). When near the Earth’s surface (below approximately 1,000 kilometers), the gravitational field remains relatively constant at about 9.8 Newtons per kilogram.

However, addressing air resistance is more complex. A basic model suggests that air resistance is proportional to the square of the velocity (v) and opposes the motion. It also factors in the air's density (ρ), the cross-sectional area (A), and a drag coefficient (C). This leads to the following relationship:

By analyzing both forces—gravity and air resistance—we can understand the motion of a falling object. If released from rest, the object initially only experiences the gravitational force, which accelerates it. As the speed increases, so does the air resistance, eventually reaching a point where both forces balance, resulting in zero net force. This constant speed is termed terminal velocity (v_t).

Yes, terminal velocity is intriguing. Now, consider a spherical rock with a density ρ_r (distinct from air density) and radius r. In this case, both the cross-sectional area (A) and mass (m) vary with changes in radius.

Inserting these variables into the terminal velocity equation leads us to a crucial conclusion:

The radius doesn't cancel out in the equation; mass grows with the cube of the radius, while area increases with the square of the radius. Hence, larger rocks possess a higher terminal velocity than smaller ones.

For instance, a typical 1 cm radius rock has a terminal velocity of around 33 meters per second (73.8 mph), assuming a density of 3,000 kg/m³ and a drag coefficient of 0.6. Now, let’s consider Asteroid M. If we estimate its size to be about 5 miles³, converting this to cubic meters and assuming a spherical shape gives us a radius of approximately 1,707 meters. This results in a staggering terminal velocity of 13,630 meters per second (30,400 mph)—quite impressive!

### Escape Velocity

Another quote from the episode mentions, “They won’t have enough power to escape Earth’s gravity.” Now, let’s explore the concept of escape velocity. What does this mean? Reverting to our rock example (since Asteroid M is essentially a massive rock), if you launch this rock upward with sufficient speed, it can ascend indefinitely without returning to Earth. The speed necessary for this is referred to as escape velocity.

Can we determine the escape velocity for Asteroid M? Absolutely! For this calculation, the Work-Energy principle is useful, stating that the work done on a system equals the change in energy. In our system comprising the Earth and the rock, we deal with two energy types: kinetic energy (K) and gravitational potential energy (U_G).

Both the Earth and the rock could possess kinetic energy, but given the massive difference in their sizes, the change in Earth’s kinetic energy is negligible. The gravitational potential energy is a characteristic of the system rather than individual objects (thus, only one U term is needed). Here, G is the universal gravitational constant, and M_E is the Earth's mass.

Imagine the rock is launched with an initial velocity v_1 while at the Earth’s surface (where r equals the radius of the Earth). With no work applied, the rock moves away to an infinite distance with zero final velocity—just barely escaping. This scenario allows us to calculate the initial velocity needed for this escape, which is the escape velocity.

It’s noteworthy that escape velocity does not depend on the mass of the rock since it cancels out in the calculation. Utilizing the Earth’s radius and mass yields an escape velocity of 11,180 meters per second. Interestingly, this value is quite close to the terminal velocity calculated previously. If we had selected a slightly smaller value for Asteroid M, the terminal velocity would surpass the escape velocity.

Of course, such a scenario is unrealistic. If the asteroid were to descend from space, it wouldn’t have the distance to reach terminal velocity. Even if it fell from low Earth orbit, it wouldn’t attain that speed. If it were in orbit, targeting its power source wouldn’t cause it to plummet.

But, it’s merely a show—a highly entertaining one at that.

The first video gives an exciting look at the X-Men attempting to prevent Asteroid M from crashing into Earth, showcasing the thrilling interplay of physics and superhero action.

The second video dives deeper into the X-Men's struggle against Asteroid M, illustrating the stakes involved in stopping a catastrophic event.