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Grade 12th passMechanics

If mass of earth decreases by 25% and it`s radius is increases by 50%,then acceleration due to gravity at its surface decreases by nearly

Profile image of Vandana Chauhan
9 Years agoGrade 12th pass
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1 Answer

Profile image of Rituraj Tiwari
5 Years ago

To determine how the acceleration due to gravity changes with a decrease in Earth's mass and an increase in its radius, we can use the formula for gravitational acceleration at the surface of a planet. The acceleration due to gravity (g) is given by the equation:

Formula for Gravitational Acceleration

The formula is:

g = G * (M / R²)

Where:

  • G is the gravitational constant (approximately 6.674 × 10-11 N(m/kg)2)
  • M is the mass of the planet (in this case, Earth)
  • R is the radius of the planet

Calculating the Changes

Let’s plug in the changes given in your question. If the mass of Earth decreases by 25%, we can express the new mass (Mnew) as:

Mnew = M * (1 - 0.25) = 0.75M

Now, if the radius increases by 50%, the new radius (Rnew) can be expressed as:

Rnew = R * (1 + 0.50) = 1.5R

Substituting Values in the Gravitational Equation

Now, we substitute these new values into the gravitational acceleration formula:

gnew = G * (Mnew / Rnew²)

Substituting the expressions we derived:

gnew = G * (0.75M / (1.5R)²)

This simplifies to:

gnew = G * (0.75M / (2.25R²))

Thus, we can further simplify this:

gnew = (G * M / R²) * (0.75 / 2.25)

Recognizing that (G * M / R²) is the original acceleration due to gravity (goriginal):

gnew = goriginal * (0.75 / 2.25)

Final Calculation

Now we calculate the fraction:

0.75 / 2.25 = 1/3

This leads us to:

gnew = goriginal * (1/3)

This means that the new acceleration due to gravity will be one-third of the original value. If we consider the average value of Earth's gravitational acceleration, which is approximately 9.81 m/s²:

gnew ≈ 9.81 m/s² * (1/3) ≈ 3.27 m/s²

Summary of Effects

In essence, if the mass of the Earth were to decrease by 25% and its radius were to increase by 50%, the acceleration due to gravity at its surface would decrease to about one-third of its original value, resulting in a new value of approximately 3.27 m/s². This significant reduction illustrates how sensitive gravitational forces are to changes in both mass and radius of a celestial body.