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Basic

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Basic

B-1. Calculating the Gravitational Force Between the Sun and Earth

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Problem:

Using Newton's law of universal gravitation \(F = Gm_1 m_2/r^2\), calculate the magnitude of the gravitational force between the Sun (mass \(M_\odot \approx 2.0 \times 10^{30}\) kg) and the Earth (mass \(M_\oplus \approx 6.0 \times 10^{24}\) kg). Take the Sun–Earth distance to be \(r \approx 1.5 \times 10^{11}\) m.

Calculation:

\[ F = \frac{G M_\odot M_\oplus}{r^2} = \frac{6.67 \times 10^{-11} \times 2.0 \times 10^{30} \times 6.0 \times 10^{24}}{(1.5 \times 10^{11})^2} \]

Numerator: \(6.67 \times 10^{-11} \times 2.0 \times 10^{30} = 1.334 \times 10^{20}\)

\(1.334 \times 10^{20} \times 6.0 \times 10^{24} = 8.00 \times 10^{44}\)

Denominator: \((1.5 \times 10^{11})^2 = 2.25 \times 10^{22}\)

\[ F = \frac{8.00 \times 10^{44}}{2.25 \times 10^{22}} \approx 3.6 \times 10^{22}\ \text{N} \]
\[ \boxed{F \approx 3.6 \times 10^{22}\ \text{N}} \]

Verification:

This corresponds to approximately \(3.6 \times 10^{18}\) tons-force. As the largest force in the solar system, this is a reasonable value for the force that keeps the Earth in its orbit.

B-2. Ratio of Gravitational to Coulomb Force Between Protons

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Problem:

Calculate the ratio of the gravitational force to the Coulomb force between two protons, \(F_{\text{grav}}/F_{\text{em}}\), by substituting specific numerical values. Use \(G \approx 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\), \(m_p \approx 1.67 \times 10^{-27}\) kg, \(e \approx 1.60 \times 10^{-19}\) C, and \(1/(4\pi\varepsilon_0) \approx 9.0 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2\).

Calculation:

\[ F_{\text{grav}} = \frac{G m_p^2}{r^2}, \qquad F_{\text{em}} = \frac{e^2}{4\pi\varepsilon_0 r^2} \]

Taking the ratio, \(r^2\) cancels:

\[ \frac{F_{\text{grav}}}{F_{\text{em}}} = \frac{G m_p^2}{e^2/(4\pi\varepsilon_0)} = \frac{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2}{9.0 \times 10^{9} \times (1.60 \times 10^{-19})^2} \]

Numerator: \(6.67 \times 10^{-11} \times 2.79 \times 10^{-54} = 1.86 \times 10^{-64}\)

Denominator: \(9.0 \times 10^{9} \times 2.56 \times 10^{-38} = 2.30 \times 10^{-28}\)

\[ \frac{F_{\text{grav}}}{F_{\text{em}}} = \frac{1.86 \times 10^{-64}}{2.30 \times 10^{-28}} \approx 8.1 \times 10^{-37} \]
\[ \boxed{\frac{F_{\text{grav}}}{F_{\text{em}}} \sim 10^{-36}} \]

Verification:

This agrees with the value \(\sim 10^{-36}\) given in the text. We have quantitatively confirmed that gravity is extraordinarily weak compared to the electromagnetic force.

B-3. Gradient of a 2D Potential

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Problem:

For the two-dimensional potential \(\Phi(x, y) = x^2 + 4y^2\), answer the following questions.

(a) Find the gradient \(\nabla\Phi = (\partial\Phi/\partial x,\;\partial\Phi/\partial y)\).

(b) Find the direction and magnitude of the gradient vector at the point \((1, 1)\).

(c) In what direction does the force \(\boldsymbol{F} = -m\nabla\Phi\) acting on an object at this point point? Explain in terms of its relationship to the equipotential lines (curves where \(\Phi = \text{const}\)).

(a) Calculating the Gradient:

\[ \frac{\partial\Phi}{\partial x} = 2x, \qquad \frac{\partial\Phi}{\partial y} = 8y \]
\[ \boxed{\nabla\Phi = (2x,\;8y)} \]

(b) Gradient Vector at Point \((1, 1)\):

Substituting \((x, y) = (1, 1)\):

\[ \nabla\Phi\big|_{(1,1)} = (2 \times 1,\;8 \times 1) = (2,\;8) \]

Magnitude:

\[ |\nabla\Phi| = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17} \approx 8.25 \]

Direction: The angle \(\theta\) measured from the positive \(x\)-axis is:

\[ \tan\theta = \frac{8}{2} = 4 \quad \Longrightarrow \quad \theta = \arctan 4 \approx 76° \]
\[ \boxed{|\nabla\Phi|_{(1,1)} = 2\sqrt{17} \approx 8.25, \quad \theta \approx 76°\ (\text{counterclockwise from the }x\text{-axis})} \]

(c) Relationship Between Force Direction and Equipotential Lines:

The force is

\[ \boldsymbol{F} = -m\nabla\Phi\big|_{(1,1)} = -m\,(2,\;8) = m\,(-2,\;-8) \]

which points in the opposite direction of the gradient vector \((2, 8)\).

The equipotential lines are ellipses defined by \(\Phi = x^2 + 4y^2 = C\) (constant). The gradient \(\nabla\Phi\) is perpendicular to the equipotential lines and points in the direction of increasing \(\Phi\) (a general property from differential geometry). Therefore, the force \(\boldsymbol{F} = -m\nabla\Phi\) is perpendicular to the equipotential lines and points in the direction of decreasing \(\Phi\) (toward the interior of the ellipse, i.e., toward the origin).

\[ \boxed{\text{The force is perpendicular to the equipotential lines (ellipses) and points in the direction of decreasing potential (toward the origin)}} \]

Verification:

  • \(\Phi(1,1) = 1 + 4 = 5\), so the equipotential line is the ellipse \(x^2 + 4y^2 = 5\). The point \((1,1)\) lies on this ellipse.
  • The normal vector to the ellipse \(x^2 + 4y^2 = C\) at point \((x_0, y_0)\) is \((2x_0, 8y_0)\), which at \((1,1)\) gives \((2, 8)\). This agrees with \(\nabla\Phi|_{(1,1)}\). This confirms that the gradient is perpendicular to the equipotential line.

B-4. Relativistic Parameter for Various Celestial Bodies

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Problem:

Calculate the criterion \(\Phi_{\text{rel}} = GM/(Rc^2)\) for the following celestial bodies.

(a) The Sun (\(M_\odot \approx 2.0 \times 10^{30}\) kg, \(R_\odot \approx 7.0 \times 10^{8}\) m)

(b) A neutron star (\(M \approx 1.4\,M_\odot\), \(R \approx 10\) km)

Express each result to 1 significant figure and discuss how reliable Newton's model is in each case.

(a) The Sun:

\[ GM_\odot = 6.67 \times 10^{-11} \times 2.0 \times 10^{30} = 1.33 \times 10^{20}\ \text{m}^3/\text{s}^2 \]
\[ R_\odot c^2 = 7.0 \times 10^{8} \times (3.0 \times 10^{8})^2 = 7.0 \times 10^{8} \times 9.0 \times 10^{16} = 6.3 \times 10^{25}\ \text{m}^3/\text{s}^2 \]
\[ \Phi_{\rm rel} = \frac{1.33 \times 10^{20}}{6.3 \times 10^{25}} \approx 2.1 \times 10^{-6} \]
\[ \boxed{\Phi_{\rm rel}(\text{Sun}) \sim 10^{-6}} \]

Newton's model is an excellent approximation near the surface of the Sun, but at the \(10^{-6}\) level of precision measurements (perihelion precession of Mercury, deflection of light), general relativistic corrections become necessary.

(b) Neutron Star:

Parameters:

  • \(M = 1.4\,M_\odot = 1.4 \times 2.0 \times 10^{30}\ \text{kg} = 2.8 \times 10^{30}\ \text{kg}\)
  • \(R = 10\ \text{km} = 1.0 \times 10^{4}\ \text{m}\)
  • \(G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\)
  • \(c = 3.0 \times 10^{8}\ \text{m/s}\)

Calculation:

Numerator: \(GM = 6.67 \times 10^{-11} \times 2.8 \times 10^{30} = 1.87 \times 10^{20}\ \text{m}^3/\text{s}^2\)

Denominator: \(Rc^2 = 1.0 \times 10^{4} \times (3.0 \times 10^{8})^2 = 9.0 \times 10^{20}\ \text{m}^3/\text{s}^2\)

Ratio:

\[ \Phi_{\rm rel} = \frac{1.87 \times 10^{20}}{9.0 \times 10^{20}} \approx 0.21 \]
\[ \boxed{\Phi_{\rm rel} \approx 0.2} \]

Discussion:

\(\Phi_{\rm rel} \approx 0.2\) is close to order unity. This means that Newton's model is insufficient for describing neutron stars, and general relativistic effects (spacetime curvature) become prominent. Calculations using Newtonian mechanics can serve as qualitative estimates, but quantitative accuracy cannot be expected.

Verification: In the table in the text, \(GM/(Rc^2)\) for a neutron star is given as \(\sim 10^{-1}\) order, and \(0.2\) is consistent with this.

B-5. Derivation of the Schwarzschild Radius

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Problem:

Using the escape velocity \(v_{\text{esc}} = \sqrt{2GM/R}\), express the radius \(R_s\) at which \(v_{\text{esc}} = c\) in terms of \(M\), \(G\), and \(c\). This \(R_s\) is called the Schwarzschild radius.

Calculation:

\[ v_{\rm esc} = \sqrt{\frac{2GM}{R}} = c \]

Squaring both sides,

\[ \frac{2GM}{R} = c^2 \]

Solving for \(R\),

\[ \boxed{R_s = \frac{2GM}{c^2}} \]

This is the Schwarzschild radius.

Verification:

Dimensional check: \([GM/c^2] = [\text{m}^3\text{s}^{-2}\cdot\text{kg}/(\text{m}^2\text{s}^{-2})] \cdot [\text{kg}]^{-1} \cdot \text{kg} = \text{m}\). Correct.

For the Sun: \(R_s = 2 \times 6.67 \times 10^{-11} \times 2.0 \times 10^{30} / (9.0 \times 10^{16}) \approx 3.0 \times 10^{3}\ \text{m} = 3\ \text{km}\). This agrees with the known value of the Sun's Schwarzschild radius.

B-6. Criterion at the Schwarzschild Radius

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Problem:

Using the Schwarzschild radius \(R_s\) obtained in Problem B-5. Derivation of the Schwarzschild Radius, calculate the criterion \(GM/(Rc^2)\) at \(R = R_s\) and verify the value corresponding to a black hole.

Calculation:

Substituting \(R_s = 2GM/c^2\),

\[ \frac{GM}{R_s c^2} = \frac{GM}{\dfrac{2GM}{c^2} \cdot c^2} = \frac{GM}{2GM} = \frac{1}{2} \]
\[ \boxed{\frac{GM}{R_s c^2} = \frac{1}{2}} \]

Discussion:

For a black hole (\(R = R_s\)), the criterion reaches \(1/2\), which is of order 1. This indicates that Newtonian mechanics completely breaks down and general relativity becomes indispensable. Phenomena that cannot be described by Newtonian mechanics arise, such as the formation of an event horizon.

Verification:

\(\Phi_{\rm rel} = 1/2\) corresponds to \(v_{\rm esc} = c\). Since \(v/c \sim \sqrt{\Phi_{\rm rel}} = 1/\sqrt{2} \approx 0.71\), the velocity is comparable to the speed of light, which is consistent with the necessity of general relativity.

B-7. Equivalence Condition of Inertial and Gravitational Mass

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Problem:

Distinguishing between inertial mass \(m_i\) and gravitational mass \(m_g\), we write the equation of motion in a uniform gravitational field \(\mathbf{g}\) as

\[ m_i\,\ddot{\boldsymbol{x}} = m_g\,\mathbf{g} \]

Find the condition, expressed in terms of the ratio of \(m_i\) to \(m_g\), for object A (\(m_i^{(A)},\; m_g^{(A)}\)) and object B (\(m_i^{(B)},\; m_g^{(B)}\)) to fall with the same acceleration.

Calculation:

The acceleration of each object follows from Newton's second law:

\[ \ddot{\boldsymbol{x}}_A = \frac{m_g^{(A)}}{m_i^{(A)}}\,\mathbf{g}, \qquad \ddot{\boldsymbol{x}}_B = \frac{m_g^{(B)}}{m_i^{(B)}}\,\mathbf{g} \]

The condition for these to be equal, \(\ddot{\boldsymbol{x}}_A = \ddot{\boldsymbol{x}}_B\), is:

\[ \frac{m_g^{(A)}}{m_i^{(A)}} = \frac{m_g^{(B)}}{m_i^{(B)}} \]
\[ \boxed{\frac{m_g^{(A)}}{m_i^{(A)}} = \frac{m_g^{(B)}}{m_i^{(B)}}} \]

That is, the condition is that the ratio of gravitational mass to inertial mass \(m_g/m_i\) is a universal constant for all objects (equal to 1 in appropriate units).

Verification:

If \(m_g/m_i = \text{const.}\), then the acceleration \(\ddot{\boldsymbol{x}} = (m_g/m_i)\,\mathbf{g}\) takes the same value regardless of the type or mass of the object. This is consistent with the result of Galileo's free-fall experiment (all objects fall with the same acceleration). Furthermore, if we set \(m_g/m_i = 1\) (the equivalence principle), we obtain \(\ddot{\boldsymbol{x}} = \mathbf{g}\), recovering the usual equation of free fall.

B-8. Basic Calculations of Partial Derivatives

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Problem:

Practice calculating partial derivatives. For the following functions, find the specified partial derivatives.

(a) For \(f(x, y) = 3x^2 y + 2y^3\), find \(\dfrac{\partial f}{\partial x}\) and \(\dfrac{\partial f}{\partial y}\).

(b) For \(g(x, y, z) = x^2 y z^3\), find \(\dfrac{\partial g}{\partial x}\), \(\dfrac{\partial g}{\partial y}\), and \(\dfrac{\partial g}{\partial z}\).

(c) For \(h(r, \theta) = r^2 \cos\theta\), find \(\dfrac{\partial h}{\partial r}\) and \(\dfrac{\partial h}{\partial \theta}\).

(a) \(f(x, y) = 3x^2 y + 2y^3\):

When differentiating partially with respect to \(x\), treat \(y\) as a constant:

\[ \frac{\partial f}{\partial x} = 6xy \]

When differentiating partially with respect to \(y\), treat \(x\) as a constant:

\[ \frac{\partial f}{\partial y} = 3x^2 + 6y^2 \]

(b) \(g(x, y, z) = x^2 y z^3\):

\[ \frac{\partial g}{\partial x} = 2xyz^3 \]
\[ \frac{\partial g}{\partial y} = x^2 z^3 \]
\[ \frac{\partial g}{\partial z} = 3x^2 y z^2 \]

(c) \(h(r, \theta) = r^2 \cos\theta\):

\[ \frac{\partial h}{\partial r} = 2r\cos\theta \]
\[ \frac{\partial h}{\partial \theta} = -r^2 \sin\theta \]

B-9. Gradient Vectors and Isotherms

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Problem:

The relationship between partial derivatives and the gradient. The function \(T(x, y) = 100 - x^2 - 4y^2\) of two variables represents the temperature distribution on a certain plane.

(a) Find \(\dfrac{\partial T}{\partial x}\) and \(\dfrac{\partial T}{\partial y}\).

(b) Calculate the values of \(\partial T/\partial x\) and \(\partial T/\partial y\) at the point \((1, 2)\), and explain in words their physical meaning ("how does the temperature change when you move slightly in the \(x\) direction?").

(c) Calculate the gradient vector \(\nabla T = (\partial T/\partial x,\;\partial T/\partial y)\) at the point \((1, 2)\), and state the direction this vector points relative to the isotherms (curves where \(T = \text{const.}\)).

(a) Partial Derivatives:

\[ \frac{\partial T}{\partial x} = -2x, \qquad \frac{\partial T}{\partial y} = -8y \]

(b) Values at the Point \((1, 2)\) and Physical Meaning:

\[ \frac{\partial T}{\partial x}\bigg|_{(1,2)} = -2(1) = -2 \]
\[ \frac{\partial T}{\partial y}\bigg|_{(1,2)} = -8(2) = -16 \]

Physical meaning:

  • \(\partial T/\partial x = -2\): Moving slightly in the \(x\) direction causes the temperature to decrease (approximately 2 degrees drop per unit distance)
  • \(\partial T/\partial y = -16\): Moving slightly in the \(y\) direction causes the temperature to decrease much more rapidly (approximately 16 degrees drop per unit distance)

We can see that the rate of change in the \(y\) direction is far steeper.

(c) Relationship Between the Gradient Vector and Isotherms:

\[ \nabla T\big|_{(1,2)} = \left(\frac{\partial T}{\partial x},\;\frac{\partial T}{\partial y}\right)\bigg|_{(1,2)} = (-2,\;-16) \]

The gradient vector is perpendicular to the isotherms (curves where \(T = \text{const.}\), which in this case are ellipses \(x^2 + 4y^2 = \text{const.}\)) and points in the direction of the steepest increase in temperature. Here it is \((-2, -16)\), pointing toward the origin—that is, in the direction where the temperature is higher.

Medium

M-1. Quantitative Evaluation of the Four Properties of Gravity

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Problem:

Based on the four fundamental properties of gravity (universality, impossibility of shielding, long-range nature, and extreme weakness), answer the following questions.

(a) Quantitatively explain why gravity is negligible at the atomic/molecular scale (\(\sim 10^{-10}\) m), using the ratio of gravitational to electromagnetic force strengths.

(b) Explain why gravity nevertheless becomes dominant at the scale of galaxy clusters (\(\sim 10^{23}\) m), clearly stating which of the four properties are essential to this explanation.

Solution:

The value of \(GM/(Rc^2)\) serves as an indicator of at what stage the transition from Newton's model to general relativity becomes necessary.

For the Earth (\(GM/(Rc^2) \sim 10^{-9}\)), general relativistic effects are extremely small but still nonzero. GPS satellite clocks experience a time drift of approximately \(45\ \mu\text{s}\) per day relative to ground clocks due to gravitational redshift, and without correction this would lead to a position error of approximately 14 km per day (this is calculated in detail in Problem A-1. Gravitational Time Dilation of GPS Satellites). Thus, even a value as small as \(10^{-9}\) makes general relativistic corrections indispensable for high-precision technology.

For the Sun (\(GM/(Rc^2) \sim 10^{-6}\)), an anomalous precession of Mercury's perihelion of approximately 43 arcseconds per century is observed that cannot be explained by Newtonian mechanics. This was the historic verification case first accurately explained by general relativity, demonstrating that effects on the order of \(10^{-6}\) are detectable through precise astronomical observations. The bending of light passing near the Sun (gravitational lensing) is also observed as an effect of comparable magnitude.

For white dwarfs (\(GM/(Rc^2) \sim 10^{-4}\)), gravitational redshift becomes clearly detectable in spectral observations, and deviations from Newtonian mechanics become more pronounced.

For neutron stars (\(GM/(Rc^2) \sim 0.1\text{–}0.2\)), Newton's model becomes quantitatively unreliable. Phenomena that cannot be described without general relativity become dominant, such as stellar structure (mass limits, relationship to the equation of state) and orbital decay due to gravitational wave emission from binary pulsars.

For black holes (\(GM/(Rc^2) = 1/2\)), Newtonian mechanics breaks down completely. Phenomena arise for which no corresponding concepts even exist within Newton's framework, such as the formation of event horizons, spacetime singularities, and gravitational wave emission during black hole mergers. In this way, as the value of \(GM/(Rc^2)\) increases, a gradual transition occurs: "Newton is sufficient" → "corrections that are tiny but detectable are needed" → "Newton is quantitatively inadequate" → "Newton breaks down completely."

M-2. Stages of the Transition from Newton to GR

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Problem:

For the celestial bodies listed in the table in the main text (Earth, Sun, white dwarf, neutron star, black hole), discuss how deviations from Newton's model become manifest as the value of \(GM/(Rc^2)\) increases, correlating them with specific observational phenomena (perihelion precession of Mercury, GPS time corrections, formation of the event horizon, etc.) (approximately 5–8 sentences).

Solution:

In Einstein's general relativity, gravity is described not as a force but as the geometric curvature of spacetime. All objects, in the absence of external forces, move along geodesics—the "straightest possible paths" in curved spacetime. The geodesic equation is determined solely by the spacetime metric and does not depend at all on the mass, composition, or internal structure of the moving object. Therefore, objects with the same initial position and initial velocity trace exactly the same trajectory regardless of their mass or material. This is precisely the experimental fact that "all objects fall with the same acceleration." In Newton's framework, this fact required the unexplained coincidence that inertial mass \(m_i\) and gravitational mass \(m_g\) happen to be equal. However, in general relativity, since gravity is incorporated into the geometry of spacetime, the equivalence \(m_i = m_g\) is naturally embedded as the starting point of the theory (the equivalence principle). It is no longer a coincidence but a fundamental principle at the very foundation of the theory.

M-3. The Relationship Between Geodesics and the Equivalence Principle

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Problem:

In Einstein's general relativity, "gravity is not a force but the curvature of spacetime." Explain in 3–5 sentences why, under this interpretation, the equivalence of inertial mass and gravitational mass (\(m_i = m_g\)) is no longer an "unexplained coincidence," from the perspective that "all objects move along geodesics in curved spacetime."

Solution Strategy: Using the concept of geodesics in general relativity, we discuss how the equivalence of inertial mass and gravitational mass is naturally explained.

Solution:

In Newtonian mechanics, the equation of motion for an object is written as \(m_i \ddot{\boldsymbol{x}} = m_g \boldsymbol{g}\), and for all objects to fall with the same acceleration \(\boldsymbol{g}\), the equality \(m_i = m_g\) must hold for all matter. However, inertial mass (the quantity that determines the response to force) and gravitational mass (the quantity that determines the source of and coupling to the gravitational field) are conceptually entirely different quantities, and within the Newtonian framework there is no theoretical justification for this equality——it can only be accepted as a mere experimental fact (a coincidence).

In Einstein's general relativity, gravity is described not as a "force" but as curvature of spacetime. Mass-energy curves spacetime, and all objects, as long as no external force acts upon them, move along geodesics——the "straightest possible paths" in spacetime——through that curved spacetime. The geodesic equation is determined solely by the spacetime metric \(g_{\mu\nu}\) and does not depend at all on the mass, composition, or internal structure of the moving object. Therefore, objects with the same initial position and initial velocity trace exactly the same trajectory, whether they are made of iron or aluminum, whether their mass is large or small. This is precisely the experimental fact that "all objects fall with the same acceleration," and the equivalence \(m_i = m_g\) is naturally incorporated as the starting point of the theory (the equivalence principle). It is no longer a "coincidence" but rather a necessary consequence of gravity being the geometry of spacetime.

\[ \boxed{\text{Since the geodesic equation does not depend on the mass or composition of the object, } m_i = m_g \text{ is automatically guaranteed by the structure of the theory.}} \]

Verification: This argument is precisely the content of the equivalence principle (weak equivalence principle), and it is experimentally supported to the precision of the Eötvös experiment (\(\eta < 10^{-13}\)). Furthermore, this can be confirmed from the fact that the mass of the object does not appear in the geodesic equation \(\dfrac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu{}_{\alpha\beta}\dfrac{dx^\alpha}{d\tau}\dfrac{dx^\beta}{d\tau} = 0\).

Advanced

A-1. Gravitational Time Dilation of GPS Satellites

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Problem:

A GPS satellite orbits in a circular orbit at an altitude of approximately \(h \approx 2.0 \times 10^{4}\) km. Take the mass of the Earth as \(M_\oplus \approx 6.0 \times 10^{24}\) kg and the radius of the Earth as \(R_\oplus \approx 6.4 \times 10^{3}\) km.

(a) Find the difference in gravitational potential between the Earth's surface and the satellite orbit: \(\Delta\Phi = \Phi(R_\oplus + h) - \Phi(R_\oplus)\).

(b) According to general relativity, between locations with a gravitational potential difference \(\Delta\Phi\), there is a relative shift in the rate at which time passes. This ratio is approximately given by

\[ \frac{\Delta\tau}{\tau} \approx \frac{\Delta\Phi}{c^2} \]

(if \(\Delta\tau/\tau > 0\), the satellite clock runs faster). Using this expression, estimate how many microseconds (\(\mu\)s) per day the satellite clock runs faster than a clock on the ground.

(c) Using the result from (b), estimate the magnitude of the position error that would accumulate per day if this time offset were not corrected.

Parameter Summary:

  • \(M_\oplus = 6.0 \times 10^{24}\ \text{kg}\)
  • \(R_\oplus = 6.4 \times 10^{3}\ \text{km} = 6.4 \times 10^{6}\ \text{m}\)
  • \(h = 2.0 \times 10^{4}\ \text{km} = 2.0 \times 10^{7}\ \text{m}\)
  • \(G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\)
  • \(c = 3.0 \times 10^{8}\ \text{m/s}\)

Orbital radius of the satellite:

\[ r_{\text{sat}} = R_\oplus + h = 6.4 \times 10^{6} + 2.0 \times 10^{7} = 2.64 \times 10^{7}\ \text{m} \]

(a) Difference in Gravitational Potential:

\[ \Delta\Phi = \Phi(R_\oplus + h) - \Phi(R_\oplus) = -\frac{GM_\oplus}{R_\oplus + h} + \frac{GM_\oplus}{R_\oplus} \]
\[ = GM_\oplus\left(\frac{1}{R_\oplus} - \frac{1}{R_\oplus + h}\right) \]

First, compute \(GM_\oplus\):

\[ GM_\oplus = 6.67 \times 10^{-11} \times 6.0 \times 10^{24} = 4.00 \times 10^{14}\ \text{m}^3/\text{s}^2 \]

Computing each term:

\[ \frac{1}{R_\oplus} = \frac{1}{6.4 \times 10^{6}} = 1.5625 \times 10^{-7}\ \text{m}^{-1} \]
\[ \frac{1}{R_\oplus + h} = \frac{1}{2.64 \times 10^{7}} = 3.788 \times 10^{-8}\ \text{m}^{-1} \]
\[ \frac{1}{R_\oplus} - \frac{1}{R_\oplus + h} = 1.5625 \times 10^{-7} - 3.788 \times 10^{-8} = 1.184 \times 10^{-7}\ \text{m}^{-1} \]
\[ \Delta\Phi = 4.00 \times 10^{14} \times 1.184 \times 10^{-7} = 4.74 \times 10^{7}\ \text{m}^2/\text{s}^2 \]
\[ \boxed{\Delta\Phi \approx 4.7 \times 10^{7}\ \text{m}^2/\text{s}^2} \]

Since \(\Delta\Phi > 0\), the gravitational potential is higher (gravity is weaker) at the satellite's position.

(b) Time Deviation per Day:

\[ \frac{\Delta\tau}{\tau} \approx \frac{\Delta\Phi}{c^2} = \frac{4.74 \times 10^{7}}{(3.0 \times 10^{8})^2} = \frac{4.74 \times 10^{7}}{9.0 \times 10^{16}} = 5.27 \times 10^{-10} \]

Since 1 day \(= 86400\) seconds, the time deviation per day is:

\[ \Delta\tau = 5.27 \times 10^{-10} \times 86400 = 4.55 \times 10^{-5}\ \text{s} = 45.5\ \mu\text{s} \]
\[ \boxed{\Delta\tau \approx 46\ \mu\text{s/day}} \]

Since \(\Delta\tau > 0\), the satellite's clock runs faster than a clock on the ground. This corresponds to time flowing more quickly due to the general relativistic gravitational redshift effect, because the satellite is in a weaker gravitational field.

(Note: In the actual GPS system, the special relativistic effect (time dilation due to the satellite's motion, approximately \(-7\ \mu\text{s/day}\)) also exists, and the net effect is approximately \(38\ \mu\text{s/day}\). In this problem, the special relativistic effect is neglected.)

(c) Position Error per Day:

If the time deviation \(\Delta\tau\) is not corrected, errors arise in distance measurements using signals propagating at the speed of light \(c\). The magnitude is:

\[ \Delta x \approx c \cdot \Delta\tau = 3.0 \times 10^{8} \times 4.55 \times 10^{-5} = 1.37 \times 10^{4}\ \text{m} \]
\[ \boxed{\Delta x \approx 14\ \text{km/day}} \]

That is, if general relativistic time corrections are not applied, position measurement errors of approximately 14 km per day accumulate in the GPS system. This is a completely unusable level of accuracy for a navigation system, demonstrating that time corrections based on general relativity are indispensable for GPS to function properly.

Verification:

  • Dimensional check: \([\Delta\Phi/c^2]\) is dimensionless. \([c \cdot \Delta\tau] = \text{m/s} \times \text{s} = \text{m}\). Correct.
  • Order-of-magnitude check: \(\Delta\Phi/c^2 \sim 5 \times 10^{-10}\) is of the same order as \(GM/(Rc^2) \sim 10^{-9}\) for the Earth, which is consistent.
  • Comparison with known values: The widely known value for the general relativistic correction in GPS is approximately \(45\ \mu\text{s/day}\) (gravitational effect only), which agrees well with our calculated value of \(46\ \mu\text{s}\). For the position error, the commonly cited rough estimate of "approximately 10 km per day" is an order-of-magnitude estimate, and it is consistent with our more precise result of approximately 14 km (the rough value corresponds to rounding to one significant figure).