How many miles must you travel to make the Earth fall the same amount?

1 answer:

4 0

Let's take the analogy of the baseball pitcher a step farther. When a baseball is thrown in a straight line, we already said that the ball would fall to Earth because of gravity and atmospheric drag. Let's pretend again that there is no atmosphere, so there is no drag to slow the baseball down. Now, let's assume that the person throwing the ball throws it so fast that as the ball falls towards the Earth, it also travels so far, before falling even a little, that the Earth's surface curves away from the ball's path.

In other words, the baseball falls as it did before, but the ball is moving so fast that the curvature of the Earth becomes a factor and the Earth "falls away" from the ball. So, theoretically, if a pitcher on a 100 foot (30.48 m) high hill threw a ball straight and fast enough,the ball would circle the Earth at exactly 100 feet and hit the pitcher in the back of the head once it circled the globe! The bad news for the person throwing the ball is that the ball will be traveling at the same speed as when they threw it, which is about 8 km/s or several times faster than a rifle bullet. This would be very bad news if it came back and hit the pitcher, but we'll get to that in a minute.

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Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

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The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .

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Answer:

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