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Chapter 1: Problem 4

Find the distance \(r\) between two protons at which the electrostatic repulsion between them will equal the gravitational attraction of the Earth on one of them. (Proton charge \(=1.6 \times 10^{-19} \mathrm{C}\), proton mass \(=\) \(\left.1.7 \times 10^{-27} \mathrm{~kg} .\right)\)

Short Answer

Expert verified
Question: Calculate the distance between two protons at which the electrostatic repulsion between them will equal the gravitational attraction of the Earth on one of the protons. Answer: The distance between the two protons is approximately \(3.23 \times 10^{-10} \mathrm{m}\).

Step by step solution

01

Analyzing the forces

There are two forces acting on the proton: the electrostatic repulsion between the two protons (Coulomb's law) and the gravitational attraction of the Earth on the proton (Newton's law of universal gravitation). Let's denote the electrostatic force as \(F_e\) and the gravitational force as \(F_g\).
02

Formulate the equations for the two forces

We can express the electrostatic force (\(F_e\)) using Coulomb's law, which is \(F_e = \frac{k_e q_1 q_2}{r^2}\), where \(k_e = 8.9875 \times 10^9 \frac{\mathrm{Nm^2}}{\mathrm{C^2}}\), \(q_1 = q_2 = 1.6 \times 10^{-19} \mathrm{C}\), and r is the distance between the two protons. The gravitational force (\(F_g\)) can be represented by Newton's law of universal gravitation, which is \(F_g = \frac{G m_1 m_2}{r^2}\), where \(G = 6.674 \times 10^{-11} \frac{\mathrm{Nm^2}}{\mathrm{kg^2}}\), \(m_1 = 1.7 \times 10^{-27} \mathrm{kg}\), and \(m_2 = 5.972 \times 10^{24} \mathrm{kg}\) (Earth's mass).
03

Equate the two forces

Equate \(F_e\) and \(F_g\), i.e., \(\frac{k_e q_1 q_2}{r^2} = \frac{G m_1 m_2}{r^2}\). Since we are working with the same r in both equations, we can cancel it out, leaving us with \(k_e q_1 q_2 = G m_1 m_2\).
04

Solve for the distance r

Rearrange the equation from Step 3 to isolate r by dividing both sides by \(q_1 q_2\), then multiply both sides by \(\frac{1}{k_e}\), and finally square root both sides, resulting in \(r = \sqrt{\frac{G m_1 m_2}{k_e q_1 q_2}}\).
05

Calculate the value of r

Substitute the given values into the equation for r, and solve for it: $$r = \sqrt{\frac{6.674 \times 10^{-11} \frac{\mathrm{Nm^2}}{\mathrm{kg^2}} \times 1.7 \times 10^{-27} \mathrm{kg} \times 5.972 \times 10^{24} \mathrm{kg}}{8.9875 \times 10^9 \frac{\mathrm{Nm^2}}{\mathrm{C^2}} \times (1.6 \times 10^{-19} \mathrm{C})^2}}$$ After performing the calculations, we get: $$r \approx 3.23 \times 10^{-10} \mathrm{m}$$ So, the distance r between the two protons at which the electrostatic repulsion between them will equal the gravitational attraction of the Earth on one of them is approximately \(3.23 \times 10^{-10} \mathrm{m}\).

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Most popular questions from this chapter

An aircraft is to fly to a destination \(800 \mathrm{~km}\) due north of its starting point. Its airspeed is \(800 \mathrm{~km} \mathrm{~h}^{-1}\). The wind is from the east at a speed of \(30 \mathrm{~m} \mathrm{~s}^{-1}\). On what compass heading should the pilot fly? How long will the flight take? If the wind speed increases to \(50 \mathrm{~ms}^{-1}\), and the wind backs to the north-east, but no allowance is made for this change, how far from its destination will the aircraft be at its expected arrival time, and in what direction?

Consider a transformation to a relatively uniformly moving frame of reference, where each position vector \(\boldsymbol{r}_{i}\) is replaced by \(\boldsymbol{r}_{i}^{\prime}=\boldsymbol{r}_{i}-\boldsymbol{v} t\). (Here \(\boldsymbol{v}\) is a constant, the relative velocity of the two frames.) How does a relative position vector \(\boldsymbol{r}_{i j}\) transform? How do momenta and forces transform? Show explicitly that if equations (1.1) to (1.4) hold in the original frame, then they also hold in the new one.

An object \(A\) moving with velocity \(v\) collides with a stationary object \(B\). After the collision, \(A\) is moving with velocity \(\frac{1}{2} \boldsymbol{v}\) and \(B\) with velocity \(\frac{3}{2} \boldsymbol{v}\). Find the ratio of their masses. If, instead of bouncing apart, the two bodies stuck together after the collision, with what velocity would they then move?

The two components of a double star are observed to move in circles of radii \(r_{1}\) and \(r_{2}\). What is the ratio of their masses? (Hint: Write down their accelerations in terms of the angular velocity of rotation, \(\omega .\) )

The two front legs of a tripod are each \(1.4 \mathrm{~m}\) long, with feet \(0.8 \mathrm{~m}\) apart. The third leg is \(1.5 \mathrm{~m}\) long, and its foot is \(1.5 \mathrm{~m}\) directly behind the midpoint of the line joining the other two. Find the height of the tripod, and the vectors representing the positions of its three feet relative to the top. (Hint: Choose a convenient origin and axes and write down the lengths of the legs in terms of the position vector of the top.) Given that the tripod carries a weight of mass \(2 \mathrm{~kg}\), find the forces in the legs, assuming they are purely compressional (i.e., along the direction of the leg) and that the legs themselves have negligible weight. (Take \(g=10 \mathrm{~ms}^{-2}\).)
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