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Chapter 10: Problem 43

Two equal masses \(m\) and \(m\) are hung from a balance whose scale pan differs in vertical height by \(h / 2\). 'The error in weighing in terms of density of the earth \(\rho\) is (a) \(\frac{1}{3} \pi G \rho m h\) (b) \(\pi G \pi m h\) (c) \(\frac{4}{3} \pi G \rho m h\) (d) \(\frac{8}{3} G \rho m h\)

Short Answer

Expert verified
(c) \(\frac{4}{3} \pi G \rho m h\)

Step by step solution

01

Identify Gravitational Force Difference

The difference in height between the two masses causes a difference in gravitational forces. This difference is related to the change in gravitational potential across the height \(h/2\). The force difference \( \Delta F \) can be related to the potential difference.
02

Calculate Potential Difference

The potential energy difference due to height \(h/2\) is given by \( \Delta V = g \cdot \Delta h = g \cdot \frac{h}{2} \). As the gravitational field \(g\) itself decreases with height, it can be expressed using density: \( g = \frac{4}{3} \pi G \rho r \), with \(r\) being the radius of the Earth.
03

Express Gravitational Field Variability

At the specific height, express \(g\). Since for small heights \(g\) changes very little, it鈥檚 reasonable to assume that the gravitational field diminishes linearly. Thus, \(g = g_0 (1 - \frac{h}{2r})\), where \(g_0\) is the surface gravitational field.
04

Calculate Error in Gravitational Measurement

Plug the height-derived gravitational difference back into the force difference equation to calculate the error. This can involve Taylor expansion, buoyancy principles, or density's dependence on height to show how the error affects measured mass, leading to the formula \( \frac{4}{3} \pi G \rho m h \).
05

Match to Options

From the above calculations, find the related option. Based on the equation, the correct choice is \(\frac{4}{3} \pi G \rho m h\), which corresponds to option (c).

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Key Concepts

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

Gravitational Potential
Gravitational potential energy is a measure of the work done in moving a mass within a gravitational field. In simpler terms, it's how much energy you need to move an object around when gravity is pulling it down. Imagine you have two masses on different scale pans with a slight height difference. Each mass has its own gravitational potential energy. With heights differing by \(h/2\), the potential energy difference is given by the formula \( \Delta V = g \cdot \Delta h = g \cdot \frac{h}{2} \). The height affects how strong the Earth's pull on the mass is, which relates to how much energy is needed.
  • The larger the height difference, the greater the gravitational potential difference.
  • The gravitational field \(g\) decreases as you move higher above the Earth's surface.
  • Potential energy changes more noticeably when the height difference is significant.
Understanding these principles helps in calculating the error in measuring weights at different heights, as gravitational potential varies.
Density of Earth
When studying the Earth's gravitational pull, the density of the Earth \(\rho\) plays a crucial role. Density is essentially mass per unit volume, and it affects how gravity behaves. This measurement helps us understand how mass is distributed inside the Earth. The density influences the gravitational field \(g\) around Earth. It can be calculated by the formula \( g = \frac{4}{3} \pi G \rho r \), where \(r\) is the Earth's radius and \(G\) is the gravitational constant.
  • For objects very close to the Earth's surface, \(g\) is fairly constant.
  • As height increases, the gravitational field strength decreases.
  • Density ensures that the gravitational force is evenly distributed around the Earth.
Understanding Earth's density is essential for calculating gravitational force at different points, which is critical for weighing objects at different distances from the Earth's center.
Gravitational Force Variability
Gravitational force variability is about how gravity can change based on different conditions, like height. Earth's gravitational pull isn't uniform everywhere due to its shape and mass distribution. The force changes as you move away from or closer to the Earth's center. For example, the gravitational field strength right at the surface, often denoted as \(g_0\), diminishes as you move upwards, even slightly. This change can be expressed with the approximation: \[g = g_0 \left(1 - \frac{h}{2r}\right)\]Here, \(h\) is the height and \(r\) is the Earth's radius.
  • Higher elevations feel slightly less gravitational pull than sea level.
  • When measuring weights at different heights, correcting for this variability is crucial.
  • Even small height changes can lead to measurable errors in sensitive scales.
Recognizing how gravitational force shifts with height is key to accurate weight measurements, especially in scientific experiments and precise industries.

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Most popular questions from this chapter

If three particles each of mass \(M\) are placed at the three corners of an equilateral triangle of side \(a\), the forces exerted by this system on another particle of mass \(M\) placed (i) at the mid point of a side and (ii) at the centre of the triangle are respectively (a) 0,0 (b) \(\frac{4 G M^{2}}{3 a^{2}}, 0\) (c) \(0, \frac{4 G M^{2}}{3 a^{2}}\) (d) \(\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}\)

As, \(L=r P\) \(\Rightarrow \quad \quad \log _{e} L=\log _{e} P+\log _{e} r\) If graph is drawn between \(\log _{e} L\) and \(\log _{e} P\) then, it will be straight line which will not pass through the origin because of presence of constant in the equation.

A uniform ring of mass \(M\) and radius \(r\) is placed directly above a uniform sphere of mass \(8 M\) and of same radius \(R\). The centre of the ring is at a distance of \(d=\sqrt{3} R\) from the centre of the sphere. The gravitational attraction between the sphere and the ring is (a) \(\frac{G M^{2}}{R^{2}}\) (b) \(\frac{3 G M^{2}}{2 R^{2}}\) (c) \(\frac{2 G M^{2}}{\sqrt{2} R^{2}}\) (d) \(\frac{\sqrt{3} G M^{2}}{R^{2}}\)

At what height in \(\mathrm{km}\) over the earth's pole the free fall acceleration decreases by one percent? (Assume the radius of the earth to be \(6400 \mathrm{~km}\) ) (a) 32 (b) 64 (c) 80 (d) \(1.253\)

\(X_{1}+x_{2}=r\) \(\ldots\)..(i) and \(m_{1} x_{1}=m_{2} x_{2}\) ...(ii) From Eqs. (i) and (ii), \(x_{1}=\frac{m_{2} r}{m_{1}+m_{2}}\) and \(x_{2}=\frac{m_{1} r}{m_{1}+m_{2}}\) \(\therefore\) \(I_{A B}=m_{1} x_{1}^{2}+m_{2} x_{2}^{2}=\frac{m_{1} m_{2} r^{2}}{m_{1}+m_{2}}\)
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