how high above the surface of the Earth would an object need to be for its weight to be half its weight on the surface?
To determine the height above the surface of the Earth at which an object's weight would be half its weight on the surface, we need to understand the relationship between an object's weight and the distance from the center of the Earth.
The weight of an object is given by the equation:
W = m * g
Where:
W is the weight of the object,
m is the mass of the object, and
g is the acceleration due to gravity.
The acceleration due to gravity, g, is approximately constant near the surface of the Earth and is approximately equal to 9.8 meters per second squared (m/s^2).
Now, let's assume the weight of the object at the surface of the Earth is W0. According to the problem, we want to find the height at which the weight is half of W0.
Since weight is directly proportional to mass, we can write:
W0 = m * g
Dividing both sides of the equation by 2, we get:
W0/2 = m * g/2
Since g/2 is still constant, let's simplify the equation:
W0/2 = m * (g/2)
Now, the weight at the new distance above the Earth's surface, let's call it h, can be calculated using the same equation:
Wh = m * (g/2)
Since the mass of the object doesn't change, we can equate the two equations:
Wh = m * (g/2) = W0/2
Now, we need to find the distance h. To do this, we can use the inverse square law for the gravitational force. The gravitational force weakens as we move farther away from the center of the Earth.
The inverse square law states:
g' = g * (R / (R + h))^2
Where:
g' is the gravitational acceleration at height h,
g is the acceleration due to gravity at the Earth's surface, and
R is the radius of the Earth (approximately 6,371 kilometers).
Plugging this expression for g' into the equation Wh = m * (g/2), we can solve for h:
Wh = m * (g/2) = m * (g * (R / (R + h))^2)
Now, we can cancel out the mass of the object:
g/2 = g * (R / (R + h))^2
Dividing both sides by g, we obtain:
1/2 = (R / (R + h))^2
Taking the square root of both sides, we get:
sqrt(1/2) = R / (R + h)
Squaring both sides of the equation, we have:
1/2 = R^2 / (R + h)^2
Now, cross-multiplying the equation, we obtain:
(R + h)^2 = 2 * R^2
Expanding and rearranging the equation, we get:
h^2 + 2 * R * h = 0
Simplifying further, we have:
h * (h + 2 * R) = 0
Since distance cannot be negative, h = 0 is not a valid solution. Thus, we conclude that:
h + 2 * R = 0
Solving for h, we find:
h = -2 * R
Since h cannot be negative in this context, we discard the negative solution. Therefore, the height above the Earth's surface at which an object's weight will be half its weight on the surface is equal to:
h = 2 * R
Substituting the known value of R ≈ 6,371 kilometers, we can calculate the height:
h ≈ 12,742 kilometers
Hence, an object would need to be approximately 12,742 kilometers above the Earth's surface for its weight to be half its weight on the surface.