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Chapter 10: Problem 56

The distance between the earth and the moon is \(3.85 \times 10^{\mathrm{B}} \mathrm{m}\). At what distance from the earth's centre, the intensity of gravitational field will be zero? The masses of earth and moon are \(5.98 \times 10^{24} \mathrm{~kg}\) and \(7.35 \times 10^{22} \mathrm{~kg}\) respectively. (a) \(3.47 \times 10^{8} \mathrm{~m}\) (b) \(0.39 \times 10^{8} \mathrm{~m}\) (c) \(1.82 \times 10^{8} \mathrm{~m}\) (d) None of these

Short Answer

Expert verified
The zero-intensity point is at approximately \(3.47 \times 10^8 \text{ m}\) from Earth's center.

Step by step solution

01

Understanding the Problem

We need to find the point where the gravitational field intensities due to Earth and the Moon cancel each other out. This is known as the point where the net gravitational force experienced by an object is zero, or the point where the gravitational field is zero.
02

Setting up the Equation

Let the distance from the Earth to the point where the intensity is zero be \( x \). Then the distance from this point to the Moon will be \( (3.85 \times 10^{8} \text{ m} - x) \). The gravitational field intensity due to the Earth at this point is given by the formula \( E_E = \frac{G M_E}{x^2} \), and due to the Moon is \( E_M = \frac{G M_M}{(3.85 \times 10^8 - x)^2} \).
03

Determining the Point Where Intensities are Zero

At the point where gravitational field intensity is zero, the intensities due to both Earth and Moon are equal: \( E_E = E_M \). Therefore, \( \frac{G M_E}{x^2} = \frac{G M_M}{(3.85 \times 10^8 - x)^2} \).
04

Solving the Equation

We can simplify by cancelling \( G \) on both sides and solving the equation: \( \frac{M_E}{x^2} = \frac{M_M}{(3.85 \times 10^8 - x)^2} \). This results in: \( M_E (3.85 \times 10^8 - x)^2 = M_M x^2 \). Substitute the given masses \( M_E = 5.98 \times 10^{24} \) kg and \( M_M = 7.35 \times 10^{22} \) kg, then solve for \( x \).
05

Calculation

Substitute the values of the masses into the equation: \((5.98 \times 10^{24}) (3.85 \times 10^8 - x)^2 = (7.35 \times 10^{22}) x^2 \). Simplifying gives a quadratic equation. Solving this equation (often by using a calculator or algebraic manipulation) will give \( x \approx 3.47 \times 10^8 \text{ m} \).
06

Verifying the Solution

Once solved, verify by substituting \( x = 3.47 \times 10^8 \) m into the original equation to ensure both sides are equal, confirming that gravitational fields at this point cancel each other.

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Key Concepts

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

Earth-Moon System
The Earth-Moon System is a fascinating astrophysical configuration where our planet Earth and its natural satellite, the Moon, interact primarily through gravitational forces. Understanding this relationship helps unveil broader cosmic dynamics. Here's a quick breakdown of what makes up this system:
  • Earth and Moon are bound together by gravity, creating a dynamic where they both rotate around a central point known as the barycenter.
  • This system directly influences tides, climatic conditions, and various biological rhythms on Earth.
  • The distance between Earth and the Moon is approximately \(3.85 \times 10^{8}\) m.
  • Their respective masses \(5.98 \times 10^{24}\) kg for Earth and \(7.35 \times 10^{22}\) kg for the Moon lead to intriguing gravitational interactions.
In our solar system, the Earth-Moon relationship serves as an excellent case study of celestial mechanics and gravitational forces, crucial for understanding larger systems like planets orbiting the Sun.
Equilibrium Point
In the context of the Earth-Moon System, an equilibrium point is where the gravitational forces exerted by the Earth and Moon balance each other out perfectly. This point is termed the Lagrange Point, although in this specific exercise, it refers to a simplified condition where gravitational field intensities equal and cancel:
  • The equilibrium point is where a small body, like a spacecraft, would experience no net gravitational force due to Earth and Moon 鈥� essentially "floating".
  • Mathematically, this point is determined where \( \frac{G M_E}{x^2} = \frac{G M_M}{(3.85 \times 10^8 - x)^2} \). The gravitational field intensities cancel each other out at a calculated distance from Earth.
  • The calculated solution from this balanced condition is approximately \(3.47 \times 10^8\) m from Earth.
Such points are crucial for space navigation and missions, as they offer stable locations for spacecraft to conserve fuel while maintaining positions within a celestial system.
Inverse Square Law
The Inverse Square Law is a fundamental principle in physics, describing how certain forces, such as gravity, weaken with distance. In gravitational terms, this principle is paramount in understanding the gravitational field intensities of celestial bodies.
  • According to the inverse square law, the intensity of a force decreases proportionally to the square of the distance from the source. For example, gravitational force \( F = \frac{G M_1 M_2}{r^2} \) demonstrates this perfectly.
  • This law governs how the Sun's light or Earth's gravitational pull lessens as one moves away.
  • In the gravitational intensity equation \( E = \frac{GM}{r^2} \), the law explains why the field intensity decreases as the distance \( r \) increases.
Hence, it becomes evident that understanding such mathematical behavior is essential for accurately determining points like the equilibrium point in the Earth-Moon system, where forces precisely counterbalance.

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Most popular questions from this chapter

Three particles each of mass \(m\) rotate in a circle of radius \(r\) with uniform angular speed \(\omega\) under their mutual gravitational attraction. If at any instant the points are on the vertex of an equilateral of side \(L\), then angular velocity \(\omega\) is (a) \(\sqrt{\frac{2 G m}{L^{3}}}\) (b) \(\sqrt{\frac{3 G m}{L^{3}}}\) (c) \(\sqrt{\frac{5 G m}{L^{3}}}\) (d) \(\sqrt{\frac{G m}{L^{3}}}\)

Two satellites \(S_{1}\) and \(S_{2}\) revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are \(1 \mathrm{~h}\) and \(8 \mathrm{~h}\) respectively. The radius of orbit of \(S_{1}\) is \(10^{4} \mathrm{~km}\). When \(S_{2}\) is closest to \(S_{1}\), the speed of \(S_{2}\) relative to \(S_{1}\) is (a) \(\pi \times 10^{4} \mathrm{kmh}^{-1}\) (b) \(2 \pi \times 10^{4} \mathrm{kmh}^{-1}\) (c) \(3 \pi \times 10^{4} \mathrm{kmh}^{-1}\) (d) \(4 \pi \times 10^{4} \mathrm{kmh}^{-1}\)

\(\mathrm{As}, I_{s}=\frac{2}{5} M R_{s}^{2}, I_{h}=\frac{2}{3} M R_{h}^{2}\) As \(\quad I_{s}=I_{h}\) \(\therefore \quad \frac{2}{5} M R_{s}^{2}=\frac{2}{3} M R_{h}^{2}\) \(\therefore\) \(\frac{R_{s}}{R_{h}}=\frac{\sqrt{5}}{\sqrt{3}}\)

As there is no external torque, angular momentum will remain constant. When the tortoise moves from \(A\) to \(C\), figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from \(C\) to \(B\), moment of inertia increases. Therefore, angular velocity decreases. If, \(M=\) mass of platform \(R=\) radius of platform \(m=\) mass of tortoise moving along the chord \(A B\) \(\mathrm{a}=\) perpendicular distance of \(O\) from \(A B\). Initial angular momentum, \(l_{1}=m R^{2}+\frac{M R^{2}}{2}\) At any time \(t\), let the tortoise reach \(D\) moving with velocity \(\underline{v}\). \(\therefore\) \(A D=v t\) As $$\begin{aligned} &A C=\sqrt{R^{2}-a^{2}} \\ &D C=A C-A D=\left(\sqrt{R^{2}-a^{2}}-v t\right) \end{aligned}$$ \(\therefore\) $$O D=r=a^{2}+\left[\sqrt{R^{2}-a^{2}}-v t\right]^{2}$$ Angular momentum at time \(t\) $$ I_{2}=m r^{2}+\frac{M R^{2}}{2} $$ As angular momentum is conserved \(\therefore\) $$l_{1} \omega_{0}=I_{2} \omega(t)$$ This shows that variation of \(\omega(t)\) with time is non-linear.

If a man weighs \(90 \mathrm{~kg}\) on the surface of earth, the height above the surface of the earth of radius \(R\), where the weight is \(30 \mathrm{~kg}\), is (a) \(0.73 R\) (b) \(R[\sqrt{3}\) (c) \(R / 3\) (d) \(\sqrt{3} R\)
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