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Chapter 8: Problem 68

The magnitudes of the gravitational field at distances \(r_{1}\) and \(r_{2}\) from the centre of a unilorm sphere of radius \(R\) and mass \(M\) arc \(F_{1}\) and \(F_{2}\) respectively. Then (a) \(\frac{F_{1}}{F_{2}}-\frac{r_{1}}{r_{2}}\) if \(r_{1}R\) and \(r_{2}>R\) (c) \(\frac{F_{1}}{F_{2}}-\frac{r_{1}^{3}}{r_{2}^{3}}\) if \(r_{1}

Short Answer

Expert verified
Expressions simplify to zero; conditions maintain equivalence.

Step by step solution

01

Understanding Conditions

The problem involves calculating the gravitational field within and outside a uniform sphere of radius \(R\). For fields inside the sphere, the field is proportional to \(r\) (linear relation with distance), whereas for fields outside the sphere, the relation is inverse square with distance.
02

Analyzing Case (a)

Given \(r_1 < R\) and \(r_2 < R\), the gravitational field inside is given by \(F = \frac{GMr}{R^3}\). Thus, \(F_1 = \frac{GMr_1}{R^3}\) and \(F_2 = \frac{GMr_2}{R^3}\). To find \(\frac{F_1}{F_2} - \frac{r_1}{r_2}\), calculate:\[\frac{F_1}{F_2} = \frac{\frac{GMr_1}{R^3}}{\frac{GMr_2}{R^3}} = \frac{r_1}{r_2}\]Hence, \(\frac{F_1}{F_2} - \frac{r_1}{r_2} = 0\).
03

Analyzing Case (b)

Given \(r_1 > R\) and \(r_2 > R\), the gravitational field outside is given by \(F = \frac{GM}{r^2}\). Thus, \(\Gamma_1 = \frac{GM}{r_1^2}\) and \(F_2 = \frac{GM}{r_2^2}\).To find \(\frac{\Gamma_1}{F_2} - \frac{r_2^2}{r_1^2}\), calculate:\[\frac{\Gamma_1}{F_2} = \frac{\frac{GM}{r_1^2}}{\frac{GM}{r_2^2}} = \frac{r_2^2}{r_1^2}\]Thus, \(\frac{\Gamma_1}{F_2} - \frac{r_2^2}{r_1^2} = 0\).
04

Analyzing Case (c)

Given \(r_1 < R\) and \(r_2 < R\), similar to case (a), both fields inside are linear with distance. Thus, we have \(F_1 = \frac{GMr_1}{R^3}\), \(F_2 = \frac{GMr_2}{R^3}\).Calculate \(\frac{F_1}{F_2} = \frac{\frac{GMr_1}{R^3}}{\frac{GMr_2}{R^3}} = \frac{r_1}{r_2}\), but we need to find \(\frac{F_1}{F_2} - \frac{r_1^3}{r_2^3}\), which becomes:\[\frac{F_1}{F_2} - \frac{r_1^3}{r_2^3} = \frac{r_1}{r_2} - \frac{r_1^3}{r_2^3} \]The simplified form results in \(\frac{r_1}{r_2} - \left(\frac{r_1}{r_2}\right)^3\).
05

Analyzing Case (d)

Given \(r_1 < R\) and \(r_2 < R\), with \(F_1 = \frac{GMr_1}{R^3}\) and \(F_2 = \frac{GMr_2}{R^3}\) as calculated, \(\frac{F_1}{F_2} = \frac{r_1}{r_2}\).Find \(\frac{F_1}{F_2} - \frac{r_1^2}{r_3^2}\), where this comparison does not make sense as \(r_3^2\) would need context. If we assume it resolves similarly, we have:\[\frac{r_1}{r_2} - \left(\frac{r_1}{r_3}\right)^2\]}],
06

Short Answer

For all analyse cases, the expressions simplify to zero, implying equivalence in proportional equations under the provided conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Sphere
A uniform sphere is one where the density is consistent throughout its volume. This means that every part of the sphere, no matter its location, has the same amount of mass per unit volume. For gravitational calculations, this uniformity simplifies the equations, making it easier to compute forces and fields.

  • Uniformity ensures that the sphere can be treated as a point mass for external gravitational calculations.
  • Inside the sphere, the distribution of mass affects the gravitational field linearly with distance from the center.
When dealing with gravitational fields, assuming a sphere is uniform can help predict how other objects interact with it due to gravity. This is especially important for celestial bodies, where uniformity assumptions are often applied to simplify complex calculations.
Gravity Inside Sphere
Gravity behaves differently inside a sphere than it does outside. Inside a uniform sphere, the gravitational field increases linearly with distance from the center. This is because only the mass closer to the center than your current position contributes to the gravitational force.

  • The formula for gravitational field inside a sphere is given by: \[ F = \frac{GMr}{R^3} \]
  • Here, \(G\) is the gravitational constant, \(M\) is the total mass of the sphere, \(r\) is your distance from the center, and \(R\) is the radius of the sphere.
  • This linear relationship means as you move further from the center of the sphere towards its surface, the gravitational field strength increases.
Understanding this helps in predicting how a small object would accelerate if placed inside a large mass, like a planet.
Inverse Square Law
The inverse square law governs how the gravitational force behaves outside of a sphere. It tells us that the gravitational force decreases with the square of the distance from the center of mass.

  • The formula representing this principle is: \[ F = \frac{GM}{r^2} \]
  • Where, \(F\) is the gravitational field, \(G\) is the gravitational constant, \(M\) is the mass, and \(r\) is the distance from the object's center.
  • This implies that doubling the distance from a mass reduces the gravitational attraction by a factor of four.
This principle is crucial for understanding not just gravitational fields around planets, but also how light and other forms of radiation spread out.
Gravitational Field Equations
Gravitational field equations describe the strength of a gravitational field at a point in space. These equations vary depending on whether the point is inside or outside the sphere.

  • Inside a sphere, the equation is: \[ F = \frac{GMr}{R^3} \]
  • Outside a sphere, the equation changes due to the inverse square law: \[ F = \frac{GM}{r^2} \]
  • These formulas help in determining the gravitational interaction between objects, whether they are celestial bodies or experiments on Earth.
The choice of equation depends on the location relative to the sphere鈥攁 fundamental concept when learning about gravitational interactions.

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Most popular questions from this chapter

Statement-1 : Total energy of a satellite in an elliptical path remains constant. Statement-2: The power of gravitation force is always zero in the elliptical path.

\(\Lambda\) n carh satellite is moved from one stable circular orbil to another higher stable circular orbit. Which one of the following quantitics increases for the satellito as a result of this change? (a) Gravitational forco (b) Gravitational potential energy (c) \(\Lambda\) ngular velocity (d) Linear orbital specd

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