The weight of a body varies inversely as the square of its distance from the center of the earth. If the radius of the earth is 4000 mi, how much would a 200 lb man weigh 1000 mi above the surface of the earth?
Well, the first step is to use the inverse square law to find the weight of the man 1000 miles above the surface of the Earth. Now, I could go into the mathematical formula and calculations, but let me put it in terms you might find a bit more amusing!
Imagine the Earth as a giant pancake, and the man as a syrup bottle. When the syrup bottle is closer to the pancake, the weight of the syrup compresses the pancake more, making it heavier. But as the syrup bottle moves up, the pancake spreads out and becomes lighter!
So, if the man weighs 200 lbs on the surface of the Earth, I'd say when he's 1000 miles up, he'll feel light enough to glide on a pancake slip 'n slide! His weight would decrease, but his joy would soar.
To solve this problem, we need to use the inverse square law. The inverse square law states that the weight of a body varies inversely as the square of its distance from the center of the earth.
Let's define the variables:
- Weight of the body: w
- Distance from the center of the earth: d
We know that the weight of the body varies inversely as the square of its distance from the center of the earth. Mathematically, this can be represented as:
w = k / d^2
Where k is the constant of proportionality.
Now, let's plug in the given values into the equation:
w = k / d^2
Weight of the man (w) = 200 lb
Distance from the center of the earth (d) = (radius of the earth + 1000 mi)
Radius of the earth = 4000 mi
So, d = 4000 + 1000 = 5000 mi
Now, we have all the values needed to determine the constant of proportionality (k).
To find k, we can rearrange the equation:
k = w * d^2
k = 200 * (5000)^2
k = 200 * 25000000
k = 5000000000
Now that we have the value of k, we can substitute it back into the equation to determine the weight of the man at a distance 1000 mi above the surface of the earth.
w = k / d^2
w = 5000000000 / (5000)^2
w = 5000000000 / 25000000
w = 200 lb
Therefore, a 200 lb man would weigh 200 lb when he is located 1000 mi above the surface of the earth.
🤣🤡
gg
The weight variation equation should be written as
w = 3200000000/d^2
before evaluating the weight for a distance of 5000 miles.
w = k(1/d^2)
when d = 4000 , w = 200
200 = k(1/16000000)
k = 200(16000000)
when d = 5000
w = 200(16000000)/5000^2 = 128