The speed at which an object would hit Earth's surface when dropped from space in a vacuum, without any air drag or friction, can be calculated using basic principles of physics.
The speed of an object in free fall near the Earth's surface can be determined using the equation:
v = √(2gh),
where v is the final velocity, g is the acceleration due to gravity (approximately 9.8 m/s² near the Earth's surface), and h is the distance the object has fallen.
In the case of an object dropped from space, we can assume that the initial height (h) is very large compared to the Earth's radius (assuming a spherical Earth). Therefore, we can consider h to be effectively infinite, and the object will continue to accelerate until it reaches the Earth's surface.
As the object falls from infinity towards the Earth's surface, its potential energy is converted into kinetic energy. Since the object starts from rest, its initial kinetic energy is zero.
The potential energy (PE) of an object at a certain height (h) can be calculated using the equation:
PE = mgh,
where m is the mass of the object and g is the acceleration due to gravity.
As the object falls towards the Earth, its potential energy is converted into kinetic energy (KE). At the point of impact with the Earth's surface, all of the potential energy is converted into kinetic energy. Therefore, we can equate the two:
mgh = (1/2)mv²,
where v is the final velocity at impact.
Simplifying the equation by canceling the mass (m) on both sides, we get:
gh = (1/2)v².
Solving for v:
v = √(2gh).
Substituting the value of g (approximately 9.8 m/s²) into the equation, we find:
v = √(2 * 9.8 * h).
Since h is effectively infinite, we can say that the object will continue to accelerate until it reaches the Earth's surface, and its velocity will approach infinity.
Therefore, in the absence of air drag or friction, the object would hit the Earth's surface at an extremely high velocity, approaching infinity.