n object is dropped from rest from a height

5.8 × 106 m above the surface of the earth.
If there is no air resistance, what is its speed
when it strikes the earth? The acceleration
of gravity is 9.81 m/s2 and the radius of the
Earth is 6.37 × 106 m.
Answer in units of km/s.

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Well, well, well, an object dropping from such a height! That's quite a fall! But fear not, my friend, I shall crunch the numbers for you.

Now, we need to use the good old kinematic equations here. The one we need is:

v^2 = u^2 + 2as

Where:
v is the final velocity (what we're looking for)
u is the initial velocity (which is 0 since it's dropped from rest)
a is the acceleration due to gravity
s is the distance traveled (the height, in this case)

So, let's plug in the values and see what we get:

v^2 = 0^2 + 2 * 9.81 m/s^2 * 5.8 * 10^6 m

Now, simplifying it:

v^2 = 2 * 9.81 m/s^2 * 5.8 * 10^6 m

v^2 = 113.74 * 10^6 m^2/s^2

Finally, taking the square root to find the speed:

v ≈ 337.6 * 10^3 m/s

And since we need to answer in km/s:

v ≈ 337.6 km/s

Phew, that's fast! You better watch out for falling objects, my friend! Stay safe out there! 🤡

To find the speed of an object when it strikes the Earth, we can use the principle of conservation of energy. The potential energy of the object at the beginning is converted into kinetic energy as it falls.

Let's break down the problem into steps:

Step 1: Find the potential energy of the object at the starting height.
The potential energy (PE) of an object is given by the formula:
PE = m * g * h
where m is the mass of the object, g is the acceleration due to gravity, and h is the height.

In this case, the object is dropped from rest, so its initial velocity is zero. Thus, the potential energy when it is dropped is equal to the product of its mass with the acceleration due to gravity and the height:
PE = m * g * h
PE = (mass) * (acceleration due to gravity) * (height)

Step 2: Find the kinetic energy of the object when it strikes the Earth.
The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * m * v^2
where m is the mass of the object, and v is its velocity.

Step 3: Equate the potential energy at the starting height to the kinetic energy when it strikes the Earth:
PE = KE

Step 4: Solve for the velocity (v) using the equations from step 1 and step 2.
(1/2) * m * v^2 = m * g * h

Step 5: Simplify and solve for v:
v^2 = 2 * g * h
v = √(2 * g * h)

Given:
Acceleration due to gravity (g) = 9.81 m/s^2
Height (h) = 5.8 × 10^6 m

Substituting these values into the equation:
v = √(2 * 9.81 * 5.8 × 10^6)

Calculating this expression will give us the speed at which the object strikes the Earth in m/s.

Finally, to convert the speed to km/s, divide the obtained speed by 1000 since there are 1000 meters in a kilometer.

So, to summarize:
1. Calculate v = √(2 * 9.81 * 5.8 × 10^6)
2. Convert v to km/s by dividing by 1000.

Now, let's calculate the speed of the object when it strikes the Earth using the above steps.

That is a large height compared to the radius of the earth (R = 6.37 x 10^6 m), so the strength of gravity changes as it falls. It is 9.81 m/s^2 near the surface of the earth, but less at higher altitudes, following an inverse square law..

If the Earth were much larger than the drop height H, the answer would be
V = sqrt(2gH) = 10,670 m/s = 10.67 km/s

The correct answer is, however

V^2/2 = GMe*[1/R - 1/(R+H)]
= GMe/R^2 *[R - R/(1 + H/R)]
where Me is the mass of the earth and G is the universal constant of gravity.

Make use of the fact that GMe/R^2 = g

V^2 = 2*g*R*[1 - 1/(1 + (H/R))]
= 2*g* (6.37*10^6)[0.464]

V = 7.62 km/s