(a) Estimate the kinetic energy of a \(100 \mathrm{m}\) diameter object, of density \(5 \mathrm{g} / \mathrm{cm}^{3},\) striking the Earth with a speed equal to the escape speed from the Earth. (b) Find one phenomenon on Earth that has a comparable energy associated with it. (c) Why is the escape speed a reasonable estimate? (Hint: Think of the speed an object would have if it fell from far away.

Escape velocity is the minimum speed an object must have to break free from the gravitational attraction of a celestial body, without further propulsion. For Earth, this speed is approximately 11.2 km/s. To calculate the escape velocity, we use the formula \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant \( \approx 6.674 \times 10^{-11} \) m\textsuperscript{3} kg\textsuperscript{-1} s\textsuperscript{-2}, \( M \) is the mass of Earth \( \approx 5.972 \times 10^{24} \) kg, and \( R \) is the radius of Earth \( \approx 6.371 \times 10^6 \) m. Objects falling from space naturally accelerate due to gravity and often reach speeds close to the escape velocity by the time they impact Earth.

Kinetic energy is the energy an object has due to its motion. It's given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. Suppose a spherical object of 100 meters diameter and a density of 5 g/cm\textsuperscript{3} (or 5000 kg/m\textsuperscript{3}) hits Earth at escape velocity (11.2 km/s). First, we find its volume \( V = \frac{4}{3}\pi(50)^3 \approx 523,598.8 \) m\textsuperscript{3}. Then, mass \( m = 5000 \times 523,598.8 \approx 2.6 \times 10^9 \) kg. Plugging these into our kinetic energy equation, we get \( KE = \frac{1}{2} \times 2.6 \times 10^9 \times (11.2 \times 10^3)^2 \approx 1.63 \times 10^{17} \) Joules.

The mass and volume of spherical objects are crucial for various calculations in physics. The volume \( V \) of a sphere is given by \( V = \frac{4}{3}\pi r^3 \), where \( r \) is the radius. For a sphere with a 100 m diameter, the radius \( r \) is 50 m, giving \( V \approx 523,598.8 \) m\textsuperscript{3}. The mass \( m \) is then calculated by multiplying the volume by its density. With a density of 5 g/cm\textsuperscript{3} (or 5000 kg/m\textsuperscript{3}), the mass is \( 5000 \times 523,598.8 \approx 2.6 \times 10^9 \) kg. Understanding these properties helps in determining the kinetic energy and potential impact force of such objects.

Natural disasters, like earthquakes, release enormous amounts of energy. To compare, the 2011 Tōhoku earthquake in Japan released about \( 1.4 \times 10^{18} \) Joules. This is in the same ballpark as the kinetic energy of our hypothetical 100 m diameter object striking Earth, calculated to be \( 1.63 \times 10^{17} \) Joules. This comparison helps to contextualize the energy involved in large-scale impacts and natural phenomena. Such analyses reveal the significant impact these events can have on Earth.