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.
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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.
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To understand the relationship between these two facts and the necessary orbit speed around Earth, we need to consider the concept of gravitational acceleration and the curvature of Earth's surface.
Fact 1 states that a freely falling object at Earth's surface drops vertically 5 meters in 1 second. This fact is related to the acceleration due to gravity, which is approximately 9.8 m/s² on Earth. This means that an object in free fall, unaffected by air resistance, will accelerate downwards at a rate of 9.8 m/s².
Fact 2 states that Earth's curvature "drops" 5 meters for every 8 kilometers of tangent. Tangent here refers to a straight line touching the surface of a curved object, in this case, Earth. It implies that if you draw a tangent line to the Earth's surface over a distance of 8 kilometers, the vertical distance between the tangent line and the surface of the Earth would be 5 meters.
Now, let's connect these facts to the necessary speed for orbiting Earth. In order for an object to orbit the Earth, it needs to have sufficient horizontal velocity to counteract the gravitational pull towards the center of the Earth. The object essentially falls towards the Earth due to gravity but maintains enough forward motion to continuously miss the surface.
The Earth's curvature affects how fast an object must travel horizontally to achieve orbit. As mentioned in Fact 2, for every 8 kilometers (or 8000 meters) of horizontal distance traveled, the vertical distance dropped due to Earth's curvature is 5 meters. This means that if an object were moving horizontally at a particular speed and for every 8000 meters it travels, it drops 5 meters due to the curvature of the Earth.
Now, if we consider Fact 1 and assume the object is in free fall while moving horizontally, it would drop 5 meters vertically in 1 second. This means that in order to continuously counteract the drop caused by the curvature of Earth, the object needs to travel far enough in that 1 second to make up for that 5-meter drop.
To achieve this, the object must be moving fast enough horizontally so that by the time 1 second elapses, it has traveled a distance far enough for the 5-meter drop to be accounted for. In other words, it needs to cover a distance equivalent to the 8-kilometer tangent mentioned in Fact 2.
Therefore, the necessary speed for orbiting Earth is related to the distance the object needs to travel horizontally within a specific time frame, taking into account Earth's curvature. By doing the necessary calculations, it is found that the object needs to travel approximately 8 kilometers horizontally in 1 second, resulting in a speed of about 8 kilometers per second or 8000 meters per second. This is the speed required to achieve a stable orbit around Earth.