Let's figure out the height above the Earth's surface where an object's weight decreases by a specific percentage. The weight of an object is the force of gravity acting on it. The force of gravity depends on the distance from the center of the Earth. As we move higher above the surface, the distance from the center of the Earth increases, and the force of gravity (and thus weight) decreases.

Understanding Weight and Gravity

The weight ($W$) of an object with mass $m$ at a certain location is given by:

\(W = m \cdot g\)

where $g$ is the acceleration due to gravity at that location.

At the surface of the Earth, let the acceleration due to gravity be $g_0$. The weight at the surface is $W_0 = m g_0$.

At a height $h$ above the surface, the acceleration due to gravity is $g_h$. The weight at this height is $W_h = m g_h$.

Relating Gravity to Height

The acceleration due to gravity at a distance $r$ from the center of the Earth is given by:

\(g = G \frac{M}{r^2}\)

where $G$ is the gravitational constant and $M$ is the mass of the Earth.

At the Earth's surface, the distance from the center is equal to the radius of the Earth, $R$. So, the acceleration due to gravity at the surface is:

\(g_0 = G \frac{M}{R^2}\)

At a height $h$ above the surface, the distance from the center is $R+h$. So, the acceleration due to gravity at height $h$ is:

\(g_h = G \frac{M}{(R+h)^2}\)

Now, we can find the ratio of $g_h$ to $g_0$:

\(\frac{g_h}{g_0} = \frac{G \frac{M}{(R+h)^2}}{G \frac{M}{R^2}} = \frac{R^2}{(R+h)^2} = \left(\frac{R}{R+h}\right)^2\)

Calculating Height for 1% Weight Reduction

The problem states that the weight of the object reduces by 1% at height $h$. This means the weight at height $h$, $W_h$, is 1% less than the weight at the surface, $W_0$.

\(W_h = W_0 - 1\% \text{ of } W_0\)

\(W_h = W_0 - 0.01 W_0\)

\(W_h = 0.99 W_0\)

Substituting the weight formulas ($W=mg$):

\(m g_h = 0.99 m g_0\)

\(g_h = 0.99 g_0\)

So, the ratio of accelerations is:

\(\frac{g_h}{g_0} = 0.99\)

We equate this to the formula relating $g_h/g_0$ to $R$ and $h$:

\(\left(\frac{R}{R+h}\right)^2 = 0.99\)

Taking the square root of both sides:

\(\frac{R}{R+h} = \sqrt{0.99}\)

Rearranging to solve for $h$:

\(R+h = \frac{R}{\sqrt{0.99}}\)

\(h = \frac{R}{\sqrt{0.99}} - R\)

\(h = R \left(\frac{1}{\sqrt{0.99}} - 1\right)\)

We are given the radius of the Earth, $R = 6400$ km.

\(h = 6400 \left(\frac{1}{\sqrt{0.99}} - 1\right)\)

\(\sqrt{0.99} \approx 0.994987\)

\(\frac{1}{\sqrt{0.99}} \approx \frac{1}{0.994987} \approx 1.005038\)

\(h \approx 6400 (1.005038 - 1)\)

\(h \approx 6400 \times 0.005038\)

\(h \approx 32.24\) km

Alternatively, since the reduction (1%) is small, we can use an approximation. For $h \ll R$, the ratio of gravity can be approximated as:

\(\frac{g_h}{g_0} = \left(1 + \frac{h}{R}\right)^{-2} \approx 1 - 2\frac{h}{R}\)

Setting this equal to 0.99:

\(1 - 2\frac{h}{R} \approx 0.99\)

\(2\frac{h}{R} \approx 1 - 0.99\)

\(2\frac{h}{R} \approx 0.01\)

\(\frac{h}{R} \approx \frac{0.01}{2} = 0.005\)

\(h \approx 0.005 R\)

Using $R = 6400$ km:

\(h \approx 0.005 \times 6400\)

\(h \approx 32\) km

Both the exact calculation and the approximation yield a value very close to 32 km, which matches one of the given options.

Therefore, the height above the surface of the Earth where the weight of an object reduces by 1% is approximately 32 km.

Revision Table: Key Concepts Reviewed

Additional Information on Gravitational Force and Weight

The force of gravity is an attractive force between any two objects with mass. For objects near the Earth's surface, this force is primarily due to the Earth's mass. Weight is essentially the magnitude of this gravitational force exerted by the Earth on an object.

• Gravity decreases with the square of the distance from the center of the attracting body. This is why gravity is weaker further away from the Earth.
• While mass is an intrinsic property of an object (it remains constant), weight is a force and depends on the local gravitational field. An object's weight would be different on the Moon or Mars compared to Earth because the gravitational acceleration is different.
• The formula \(g_h = G \frac{M}{(R+h)^2}\) shows that gravity depends on the mass of the Earth ($M$) and the distance from its center ($R+h$).