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state-variation-with-depth

🚀 Gravitation is a fundamental force that attracts two bodies towards each other. The acceleration due to gravity (g) is the rate at which an object accelerates towards the Earth due to this gravitational force. It varies with depth below the Earth's surface. As we go deeper into the Earth, the value of g decreases because the mass of the Earth above the object exerts a gravitational pull that reduces the net gravitational force acting on it. This concept is crucial for understanding how gravity behaves in different environments, especially in physics and engineering applications.

Theory Explanation

Understanding Gravitational Force

The gravitational force between two masses is given by Newton's law of gravitation, which states that every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

\[ F = G \frac{m_1 m_2}{r^2} \]
Acceleration Due to Gravity (g)

The acceleration due to gravity at the surface of the Earth is approximately 9.81 m/s². This value can be derived from the gravitational force acting on a mass at the Earth's surface. The formula for g is given by g = G * M / R², where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.

\[ g = G \frac{M}{R^2} \]
Variation of g with Depth

As we move below the Earth's surface, the value of g decreases. This is because the effective mass of the Earth that contributes to the gravitational pull decreases with depth. The formula for g at a depth d is given by g_d = g (1 - d/R), where g is the acceleration due to gravity at the surface, d is the depth, and R is the radius of the Earth.

\[ g_d = g \left(1 - \frac{d}{R}\right) \]

Key Points

  • 🎯 Gravitational force is universal and acts between all masses.
  • 🎯 The acceleration due to gravity (g) is approximately 9.81 m/s² at the Earth's surface.
  • 🎯 g decreases as we go deeper into the Earth due to the reduction in effective mass above the object.

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Examples:💡

Calculate the acceleration due to gravity at a depth of 1000 km below the Earth's surface.

Solution:

Step 1: Given that the radius of the Earth (R) is approximately 6371 km and g at the surface is 9.81 m/s², we can use the formula g_d = g (1 - d/R). Here, d = 1000 km.

\[ g_d = 9.81 \left(1 - \frac{1000}{6371}\right) \]

Step 2: Calculate the fraction: 1000/6371 ≈ 0.1564.

\[ g_d = 9.81 \left(1 - 0.1564\right) \]

Step 3: Now calculate g_d: g_d = 9.81 * 0.8436 ≈ 8.28 m/s².

\[ g_d \approx 8.28 \text{ m/s}^2 \]

Determine the value of g at a depth of 2000 km.

Solution:

Step 1: Using the same formula g_d = g (1 - d/R), where d = 2000 km.

\[ g_d = 9.81 \left(1 - \frac{2000}{6371}\right) \]

Step 2: Calculate the fraction: 2000/6371 ≈ 0.3130.

\[ g_d = 9.81 \left(1 - 0.3130\right) \]

Step 3: Now calculate g_d: g_d = 9.81 * 0.6870 ≈ 6.75 m/s².

\[ g_d \approx 6.75 \text{ m/s}^2 \]

Common Mistakes

  • Mistake: Students often forget to convert units when calculating depth in kilometers and radius in meters.

    Correction: Always ensure that all units are consistent, converting kilometers to meters if necessary.

  • Mistake: Some students assume that g remains constant at all depths, which is incorrect.

    Correction: Remember that g decreases with depth; use the appropriate formula to calculate g at different depths.

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