The weight of an object varies inversely as the square of its distance from the centre of the earth. The weight of an aeroplane is observed to be only 4/9 of its weight on earth. How far is the aeroplane from the surface of the earth. Take the radius of the earth to be 6400km
W=k/d^2.
Distance = r+R.
Distance from the centre, r=0
W=k/R×R
K =W×R×R
W=w×R×R/(r+R)^2
4W/9=(W×R×R)/(r+R)^2
4W(r+R)^2= 9×W×R×R
4(r+R)^2=9×R×R
(r+6400)^2= 9×6400×6400/4
r+6400= 9600
r = 9600-6400
r=3200km
aeroplane ... or space capsule?
inverse square root of 3/2
(3/2 * 6400) - 6400
woof ... not fully awake yet
inverse square root of 4/9 is 3/2
(3/2 * 6400) - 6400 = ?
Well, well, well, if it isn't one of those inverse square law problems. Time for some mathematical clowning around!
Let's start with some clown logic. If the weight of the aeroplane is only 4/9 of its weight on Earth, we can say that the weight (W) is inversely proportional to the square of the distance (d):
W ∝ 1/d^2
Now, let's add some clown math to the equation. Since we know the weight on Earth (W_earth) and we want to find the distance (d), we can write the equation as:
W / W_earth = (1/d^2) / (1/r^2)
Here, r represents the radius of the Earth (6400km).
Now, let's get rid of these pesky fractions and multiply both sides of the equation by (r^2):
W * r^2 = W_earth * d^2
Since we know the weight on Earth is W_earth and the weight of the aeroplane is 4/9 of that, we can substitute those values:
(4/9) * W_earth * r^2 = W_earth * d^2
Now, let's cancel out W_earth on both sides:
(4/9) * r^2 = d^2
Finally, let's take the square root of both sides to find the distance (d):
d = √((4/9) * r^2)
Plugging in the value of r (6400km) in our clown calculator:
d ≈ √((4/9) * (6400)^2)
Calculating...
d ≈ √((4/9) * 40960000)
d ≈ √(18222222.222)
d ≈ 4269.07 km
So, according to my clown calculations, the aeroplane is approximately 4269.07 km from the surface of the Earth.
To solve this problem, let's assume that the weight of the airplane on the surface of the Earth is W, and its weight at a distance x from the surface of the Earth is W'.
According to the given information, the weight of the airplane (W') is observed to be 4/9 of its weight on the surface of the Earth (W). Mathematically, this can be expressed as:
W' = (4/9)W
We also know that weight varies inversely as the square of the distance from the center of the Earth. This means that the weight (W') is inversely proportional to the square of the distance (x) from the surface of the Earth.
Mathematically, this can be expressed as:
W' = k/x^2
Where k is a constant of variation.
So, we have two equations:
W' = (4/9)W
W' = k/x^2
We can now equate these two expressions for W':
(4/9)W = k/x^2
Now, let's consider the weight of the airplane on the surface of the Earth (W). We know that weight is given by:
W = mg
Where m is the mass of the airplane and g is the acceleration due to gravity on the surface of the Earth.
Since we are dealing with weight as a proportion, we can ignore the mass and use g as a constant. The value of g is approximately 9.8 m/s^2.
W = (mass of airplane) * g
We can now substitute W into the equation (4/9)W = k/x^2:
(4/9) * (mass of airplane) * g = k/x^2
Now, let's consider the value of k. The value of k can be determined by plugging in the given values for weight and distance at a specific point. We'll use the weight on the Earth's surface and the radius of the Earth.
W = (mass of airplane) * g
W = (mass of airplane) * 9.8 (since g = 9.8 m/s^2)
From the question, we know that W' = (4/9)W, so let's plug this into the equation for W':
(4/9)W = k/x^2
(4/9) * (mass of airplane) * 9.8 = k/x^2
Simplifying, we get:
(4/9) * 9.8 = k/x^2
4 * 9.8 = k/x^2
k = (4 * 9.8) / x^2
Now, we can substitute this value of k back into our equation:
(4/9) * (mass of airplane) * 9.8 = (4 * 9.8) / x^2 * 1/x^2
Cancelling out the '9.8' and rearranging, we get:
mass of airplane = (4x^2) / 9
Now, we will substitute the value of the airplane's weight (W') in terms of its mass (W' = (4/9)W) into this equation:
(4/9)W = (4x^2) / 9
Cancel out the '4/9' on both sides:
W = x^2
We know that weight on the surface of the Earth (W) is proportional to the square of the distance (x). Thus, W and x^2 represent the same concept. Therefore, we can conclude that:
x^2 = r^2
Where r is the radius of the Earth. As given in the question, the radius of the Earth is 6400 km.
Therefore, x^2 = (6400^2) km^2
x = 6400 km
To answer the question, the aeroplane is located at a distance of 6400 km from the surface of the Earth.