Two facts: A freely falling object at Earth’s surface drops vertically 5 m in 1 s. Earth’s curvature “drops” 5 m for each 8-km tangent. Discuss how these two facts relate to the 8-km/s speed necessary to orbit Earth.

To understand the relationship between the two given facts, let's break them down.

Fact 1: A freely falling object at Earth’s surface drops vertically 5 m in 1 s.
This fact tells us that an object in free fall (experiencing only the force of gravity) will fall 5 meters in the span of 1 second. This implies that the object's average velocity during that 1-second interval is 5 m/s downward.

Fact 2: Earth’s curvature "drops" 5 m for each 8-km tangent.
This fact refers to the curvature of the Earth's surface. It means that for every straight line section of 8 kilometers along the Earth's surface (tangent), the vertical distance drops by 5 meters. So, if you were to draw a tangent line on the Earth's surface that extends for 8 kilometers, there would be a 5-meter drop in the vertical distance.

Now, let's discuss how these two facts relate to the speed necessary to orbit Earth.

For an object to remain in orbit around the Earth, it needs to have sufficient horizontal velocity to maintain a curved trajectory where it constantly falls towards the Earth while also moving forward enough to not crash into it. This balance between falling towards the Earth and moving forward is what keeps an object in orbit.

When an object falls, it follows a curved path due to the Earth's gravitational pull. The rate at which it falls depends on its velocity and the curvature of the Earth's surface. In order to stay in orbit, the object needs to be moving fast enough horizontally so that it continually "misses" the Earth as it falls towards it.

Now, let's consider a hypothetical object at the point where the Earth's curvature "drops" by 5 meters for every 8-kilometer tangent. If we assume that the object is at a height of 5 meters above the Earth's surface at this point and has 0 horizontal velocity, it would fall straight down and touch the ground in 1 second, in accordance with Fact 1.

To achieve orbit at this specific point, the object would need to have a horizontal velocity that matches the curvature of the Earth's surface. Since the Earth's curvature drops by 5 meters for every 8-kilometer tangent, this means that for every second the object falls vertically 5 meters, it needs to move forward by 8 kilometers.

Now, if we calculate the average velocity required for the object to move forward by 8 kilometers in 1 second, we get 8 km/s. This speed is necessary for the object to achieve a circular orbit at this specific point on the Earth's surface.

It's important to note that this specific relationship between the given facts and the orbital speed of 8 km/s only applies to the scenario mentioned. In reality, the orbital speed around Earth varies depending on the altitude and specific orbit desired.