Question

If the radius of earth is \(\mathrm{R}\) then height \({ }^{\prime} \mathrm{h}\) ' at which value of ' \(\mathrm{g}\) ' becomes one-fourth is (A) \(\mathrm{R} / 4\) (B) \(3 \mathrm{R} / 4\) (C) \(\mathrm{R}\) (D) \(\mathrm{R} / 8\)

Short answer

The height at which the gravitational acceleration (g) becomes one-fourth of its value on Earth's surface is \(h = R\). So, the correct option is (C).

Step-by-step solution

1. Formula for gravitational force

We know that the formula for gravitational force is given by: \[F = G \frac{m \cdot M}{r^2}\] Where: * F is the gravitational force * G is the gravitational constant * m and M are the two masses * r is the distance between the centers of masses

2. Calculate gravitational acceleration on Earth's surface

The acceleration due to gravity on the Earth's surface, g, is related to the gravitational force by: \( F = m \cdot g\) Using the formula from the first step, we can find the gravitational acceleration on Earth's surface: \(g = \frac{GM_\oplus}{R^2}\) Here, Mₑ represents the mass of Earth.

3. Find the value of g which is one-fourth

Given that the value of g becomes one-fourth at height h, the new value of g can be represented as: \(g_{new} = \frac{1}{4} \cdot g = \frac{1}{4}\frac{GM_\oplus}{R^2}\)

4. Calculate g at height h

We need to find the value of g at height h. The formula for g at height h can be given by: \(g_{h} = \frac{GM_\oplus}{(R+h)^2}\)

5. Equate g_new with g_h and solve for h

Now, equate g_new with g_h: \(\frac{1}{4}\frac{GM_\oplus}{R^2} = \frac{GM_\oplus}{(R+h)^2}\) Solve for h: \(\frac{1}{4}R^2 = (R+h)^2\) \(R^2 = 4(R+h)^2\) \(R^2 = 4R^2 + 8Rh + 4h^2\) \(-3R^2 = 8Rh + 4h^2\) \(-\frac{3}{4}R^2 = 2h^2 + 4Rh\) \(-\frac{3}{8}R^2 = h^2 + 2Rh\) Now, we have a quadratic equation in terms of h: \(h^2 + 2Rh -\frac{3}{8}R^2 = 0\)

6. Solve the quadratic equation

Now, solve this quadratic equation for h using the quadratic formula: \(h = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\) In this case, we have \(a = 1\), \(b = 2R\), and \(c = -\frac{3}{8}R^2\). Plug in the values and solve for h: \(h = \frac{-2R\pm\sqrt{(2R)^2 - 4(\frac{3}{8}R^2)}}{2}\) \(h = -R\pm\sqrt{R^2}\) One solution is \(h = -R\), but the height cannot be negative, so we can discard this solution. The other solution is \(h = R\).

Conclusion

Therefore, the height at which the gravitational acceleration (g) becomes one-fourth of its value on Earth's surface is \(h = R\). So, the correct option is (C).

Key concepts

Gravitational Force

Gravitational force is the attractive force acting between two masses due to their mass. The basic formula governing this force is expressed as:
\[F = G \frac{m \cdot M}{r^2}\]
Here,

• \(F\) stands for the gravitational force between two objects.
• \(G\) is the gravitational constant, which remains the same throughout the universe.
• \(m\) and \(M\) are the masses of the two objects, respectively.
• \(r\) is the distance between the centers of the two masses.

This formula signifies that the force is inversely proportional to the square of the distance between two masses, meaning the force decreases as the distance increases. At the Earth's surface, this force generates the acceleration known as gravity (\(g\)). Understanding gravitational force helps us comprehend how the planets orbit the Sun and how objects fall to the ground on Earth.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two, and it takes the general form:
\[ax^2 + bx + c = 0\]
In the context of the exercise, we encountered a quadratic equation while calculating the height at which gravitational acceleration becomes one-fourth of its value at Earth's surface. The specific equation derived was:
\[h^2 + 2Rh -\frac{3}{8}R^2 = 0\]
To solve this equation, the quadratic formula is used:
\[h = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\]

• \(a\) represents the coefficient of \(h^2\).
• \(b\) is the coefficient of \(h\).
• \(c\) is the constant term.

By substituting the respective values into the formula, we can find the height \(h\). Quadratic equations are essential in mathematics as they frequently appear in various areas, including physics and engineering. They enable us to solve a broad range of practical problems, such as finding maximums, minimums, and zero points.

Earth's Radius

The Earth's radius is a fundamental value when performing calculations involving gravitational force and acceleration. It is approximately 6,371 kilometers or about 3,959 miles. In our exercise, the Earth's radius is noted as \(R\), which serves as a reference point for both the surface level gravitational acceleration and calculations involving height above the surface.
Understanding Earth's radius is important because:

• It helps determine the surface gravity \(g\), which can be calculated using the formula \(g = \frac{GM_\oplus}{R^2}\).
• It acts as a benchmark for evaluating changes in gravity as you move away from the Earth's surface. For example, moving to a height \(h\) above the surface changes the effective distance to \((R+h)\).
• It assists in calculating geophysical and astronomical phenomena, such as the Earth's orbit and satellite positioning.

Overall, Earth's radius is a crucial parameter for many scientific calculations, allowing us to understand both terrestrial and cosmic movements.

Acceleration Due to Gravity

Acceleration due to gravity, often denoted as \(g\), is the rate at which an object accelerates when falling freely towards the Earth, under the influence of the Earth's gravitational force. At sea level, this acceleration is approximately 9.81 meters per second squared (m/s²).
It's represented by the equation:
\[g = \frac{GM_\oplus}{R^2}\]
This equation shows that:

• \(G\) is the gravitational constant.
• \(M_\oplus\) is the mass of the Earth.
• \(R\) is the Earth’s radius.

In the given exercise, we explored how \(g\) is reduced to one-fourth at a distance \(h\) above Earth's surface. The reduced gravitational acceleration at height \(h\) is:
\[g_{h} = \frac{GM_\oplus}{(R+h)^2}\]
As \(h\) increases, \(g\) decreases due to the greater distance from the Earth's centre. Understanding acceleration due to gravity is vital in many fields, such as engineering, physics, and astrophysics, as it affects everything from falling objects to orbits of satellites.