THIS IS A QUESTION IN MY PHYSICS BOOK & I REALLY REALLY NEED HELP. I AM 48YRS OF AGE JUST DECIDED TO GO BACK TO SCHOOL AND HONESTLY KNOW NOTHING ABOUT PHYSICS.

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.

an object in orbit is constantly falling toward the earth. However, its lateral motion takes it away from falling directly down, and by the time it has moved sideways, the curvature of the earth makes the surface farther away. At the right speed, the surface recedes at the same speed the object is falling, so it never hits.

To understand how these two facts relate to the 8-km/s speed necessary to orbit Earth, we need to break down the problem step by step.

First, let's focus on the freely falling object at Earth's surface. We are given that the object drops vertically 5 m in 1 s. This fact tells us about the acceleration due to gravity. In the absence of air resistance, all objects near the Earth's surface experience a constant acceleration of approximately 9.8 m/s^2 downward.

Now, let's consider Earth's curvature. We are told that Earth's curvature "drops" 5 m for each 8-km tangent. This refers to the fact that the surface of the Earth is not perfectly flat but curved. For every 8 kilometers of distance travelled along the Earth's surface, there is a vertical drop of 5 meters due to the curvature.

To relate these two facts to the 8-km/s speed necessary to orbit Earth, we need to consider the concept of orbital velocity. Orbital velocity is the minimum velocity required for an object to remain in a stable orbit around the Earth.

When an object is in orbit, it is continuously falling towards the Earth due to gravity. However, it is also moving forward with enough speed that it constantly "misses" the Earth and keeps falling towards it but never actually reaches the surface.

Now, let's bring all this information together. We know that for a freely falling object at Earth's surface, it drops vertically 5 m in 1 s. Taking into account the acceleration due to gravity, we can calculate the initial velocity (v) required for the object to fall 5 m in 1 s, which is approximately 9.8 m/s.

Next, we consider the Earth's curvature. For each 8 kilometers of distance travelled along the Earth's surface, there is a vertical drop of 5 meters. This means that if an object maintains an altitude of 8 kilometers above the Earth's surface and moves horizontally, it will experience a drop of 5 meters due to the Earth's curvature.

To remain in orbit, an object needs to continuously fall towards the Earth while also moving horizontally fast enough to maintain its altitude. This means that the horizontal speed required to orbit the Earth should compensate for the 5-meter drop per 8 kilometers travelled due to Earth's curvature.

Since the time it takes for the object to travel 8 kilometers horizontally is 8 km divided by the horizontal speed (v), which we are trying to find, we can use the fact that the curvature "drops" 5 meters for each 8-km tangent to set up the equation:

5 m = (8 km / v) x 5 m

Solving this equation will give us the value of v, the necessary horizontal speed to compensate for the Earth's curvature. Substituting the values, we can solve for v:

5 m = (8 km / v) x 5 m
5 m = (8,000 m / v) x 5 m
5 = (8,000 / v)

From here, we can solve for v by dividing both sides of the equation by 5:

5 / 5 = (8,000 / v) / 5
1 = (8,000 / v)

Rearranging the equation, we have:

v = 8,000

Therefore, the horizontal speed required to orbit the Earth is approximately 8,000 m/s, or 8 km/s.

In conclusion, these two facts about the freely falling object and Earth's curvature are related to the 8-km/s speed necessary to orbit Earth. The fact that a freely falling object at Earth's surface drops vertically 5 m in 1 s gives us an understanding of the acceleration due to gravity, while the fact that Earth's curvature "drops" 5 m for each 8-km tangent relates to the necessary horizontal speed to maintain altitude and compensate for the Earth's curvature when orbiting the Earth.