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Chapter 2: Problem 5

Write down the potential energy function corresponding to the force \(-G M m / x^{2}\) experienced by a particle of mass \(m\) at a distance \(x\) from a planet of mass \(M(\gg m)\). The particle is released from rest at a distance \(a\) from the centre of the planet, whose radius is \(R\), and falls under gravity. What is its speed when it strikes the surface? Evaluate this speed for a particle falling to the Earth from \(a=2 R\). (Use \(R=6400\) \(\mathrm{km}\), and \(G M / R^{2}=g=10 \mathrm{~ms}^{-2}\).)

Short Answer

Expert verified
Question: Calculate the speed of a particle released from rest at a distance 2R from the center of a planet, where R is the planet's radius, when it strikes the surface of the planet. Answer: The speed of the particle when it strikes the surface of the planet is approximately \(7.9 \times 10^3 \,\mathrm{ms^{-1}}\).

Step by step solution

01

Write down the potential energy function

Start by the given gravitational force: \(F = -\frac{G M m}{x^2}\) Recall that the change in potential energy can be found by taking the negative integral of the force with respect to distance x: \(U(x) = - \int F \,\mathrm{d}x\) Integrate the gravitational force to find the potential energy function: \(U(x) = - \int -\frac{G M m}{x^2} \,\mathrm{d}x = G M m \int \frac{1}{x^2} \,\mathrm{d}x\) As a result, we obtain: \(U(x) = -\frac{G M m}{x} + C\), where C is the integration constant that depends on the reference point for potential energy.
02

Apply conservation of mechanical energy

The initial mechanical energy is: \(E_a = U(a) + \frac{1}{2}m v^2\) Since the particle is released from rest, its initial kinetic energy is zero, and: \(E_a = U(a) = -\frac{G M m}{a} + C\) The mechanical energy when the particle reaches the planet's surface (at x = R) is: \(E_R = U(R) + \frac{1}{2}m v_R^2\) Due to the conservation of mechanical energy, \(E_a = E_R\). Therefore, we have: \(-\frac{G M m}{a} + C = -\frac{G M m}{R} +\frac{1}{2}m v_R^2\) Next, isolate \(v_R\) and solve for the speed: \(v_R^2 = 2 \left(\frac{G M m}{R}-\frac{G M m}{a}\right)\) \(v_R = \sqrt{2 \left(\frac{G M m}{R}-\frac{G M m}{a}\right)}\)
03

Calculate the speed of the particle falling to Earth

Now, plug in the given values and constants: - \(a = 2R\) - \(R = 6400\) km \(= 6.4 \times 10^6\) m - \(\frac{GM}{R^2} = g = 10\) \(\mathrm{ms^{-2}}\) We can find \(GM\) using the expression above: \(GM = gR^2 = (10 \,\mathrm{ms^{-2}})(6.4 \times 10^6 \,\mathrm{m})^2\) Substitute the values in the speed equation: \(v_R = \sqrt{2 \left(\frac{GM}{6.4 \times 10^6 \,\mathrm{m}}-\frac{GM}{2(6.4 \times 10^6 \,\mathrm{m})}\right)}\) Calculate the speed of the particle: \(v_R \approx 7.9 \times 10^3 \,\mathrm{ms^{-1}}\) Hence, the speed of the particle when it strikes Earth's surface is approximately \(7.9 \times 10^3 \,\mathrm{ms^{-1}}\).

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Most popular questions from this chapter

*The potential energy function of a particle of mass \(m\) is \(V=c x /\left(x^{2}+\right.\) \(\left.a^{2}\right)\), where \(c\) and \(a\) are positive constants. Sketch \(V\) as a function of \(x\). Find the position of stable equilibrium, and the period of small oscillations about it. Given that the particle starts from this point with velocity \(v\), find the ranges of values of \(v\) for which it (a) oscillates, (b) escapes to \(-\infty\), and \((c)\) escapes to \(+\infty\).

A particle of mass \(m\) moves (in the region \(x>0\) ) under a force \(F=\) \(-k x+c / x\), where \(k\) and \(c\) are positive constants. Find the corresponding potential energy function. Determine the position of equilibrium, and the frequency of small oscillations about it.

A particle moving under a conservative force oscillates between \(x_{1}\) and \(x_{2}\). Show that the period of oscillation is $$ \tau=2 \int_{x_{1}}^{x_{2}} \sqrt{\frac{m}{2\left[V\left(x_{2}\right)-V(x)\right]}} \mathrm{d} x $$ In particular, if \(V=\frac{1}{2} m \omega_{0}^{2}\left(x^{2}-b x^{4}\right)\), show that the period for oscillations of amplitude \(a\) is $$ \tau=\frac{2}{\omega_{0}} \int_{-a}^{a} \frac{\mathrm{d} x}{\sqrt{a^{2}-x^{2}} \sqrt{1-b\left(a^{2}+x^{2}\right)}} $$ Using the binomial theorem to expand in powers of \(b\), and the substitution \(x=a \sin \theta\), show that for small amplitude the period is approximately \(\tau \approx 2 \pi\left(1+\frac{3}{4} b a^{2}\right) / \omega_{0}\).

A particle falling under gravity is subject to a retarding force proportional to its velocity. Find its position as a function of time, if it starts from rest, and show that it will eventually reach a terminal velocity. [The equation of motion can be integrated once to give, with a suitable choice of origin and definition of \(\gamma\) (differing from \((2.28)\) by a factor of 2), \(\dot{z}+\gamma z=-g t\). To integrate again, use an integrating factor, i.e. a function \(f(t)\) such that when the equation is multiplied by \(f(t)\) the left- hand side becomes an exact derivative, in fact the derivative of \(z f\). The final stage requires an integration by parts.]

A particle moves vertically under gravity and a retarding force proportional to the square of its velocity. (This is more appropriate than a linear relation for larger particles - see Problem 8.10.) If \(v\) is its upward or downward speed, show that \(\dot{v}=\mp g-k v^{2}\), respectively, where \(k\) is a constant. If the particle is moving upwards, show that its position at time \(t\) is given by \(z=z_{0}+(1 / k) \ln \cos \left[\sqrt{g k}\left(t_{0}-t\right)\right]\), where \(z_{0}\) and \(t_{0}\) are integration constants. If its initial velocity at \(t=0\) is \(u\), find the time at which it comes instantaneously to rest, and its height then. [Note that \(\ln\) always denotes the natural logarithm: \(\ln x \equiv \log _{\mathrm{e}} x\). You may find the identity \(\ln \cos x=-\frac{1}{2} \ln \left(1+\tan ^{2} x\right)\) useful.]
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