Question

The height vertically above the Earth's surface at which the acceleration due to gravity becomes 1 percent of its value at the surface is

• A. 8R
• B. 9R
• C. 10R
• D. 20R

Hint:

The acceleration due to gravity decreases with the square of the distance from the center of the Earth. To find the height where gravity is a specific fraction of its surface value, use the formula \( g_h = \frac{g_0}{(1 + \frac{h}{R})^2} \).

Answer 1

The Correct Option is B

The acceleration due to gravity at a height \( h \) above the Earth's surface is given by the formula: \[ g_h = \frac{g_0}{(1 + \frac{h}{R})^2} \] where:
- \( g_h \) is the acceleration due to gravity at height \( h \),
- \( g_0 \) is the acceleration due to gravity at the surface of the Earth,
- \( R \) is the radius of the Earth,
- \( h \) is the height above the Earth's surface. We are given that \( g_h = 0.01 g_0 \) (1 percent of the surface value). Substituting this into the equation: \[ 0.01 g_0 = \frac{g_0}{(1 + \frac{h}{R})^2} \] Dividing both sides by \( g_0 \): \[ 0.01 = \frac{1}{(1 + \frac{h}{R})^2} \] Taking the reciprocal of both sides: \[ 100 = (1 + \frac{h}{R})^2 \] Taking the square root of both sides: \[ 10 = 1 + \frac{h}{R} \] Solving for \( h \): \[ \frac{h}{R} = 9 \]
Thus, the height is: \[ h = 9R \] Therefore, the correct answer is: \[ \text{(B) } 9R \]

Answer 2

To solve this problem, we need to find the height above the Earth's surface where the acceleration due to gravity is 1% of its value on the surface. The acceleration due to gravity at the Earth's surface is given by \( g \), and it decreases with height. The formula for gravity at height \( h \) is:

\[ g_h = \frac{g}{\left(1+\frac{h}{R}\right)^2} \]

where \( R \) is the radius of the Earth. At the desired height, gravity becomes 1% of its original value:

\[ \frac{g}{100} = \frac{g}{\left(1+\frac{h}{R}\right)^2} \]

Cancelling \( g \) from both sides and solving for \( h \):

\[ \frac{1}{100} = \frac{1}{\left(1+\frac{h}{R}\right)^2} \]

\[ 100 = \left(1+\frac{h}{R}\right)^2 \]

Taking the square root of both sides:

\[ 10 = 1+\frac{h}{R} \]

Solving for \( h \):

\[ \frac{h}{R} = 9 \]

Thus,

\[ h = 9R \]

The height where gravity is 1% of its surface value is \( 9R \).