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Chapter 3: Problem 79

In the late twentieth century the Pioneer 10 and 11 space probes were passing through the Kuiper belt, a region beyond Neptune that contains dust, ice and other small bodies, and which is thought to be the origin of some comets. Pioneer 10 had a mass of \(240 \mathrm{~kg}\), was travelling at a velocity of \(12.2 \mathrm{~km} / \mathrm{s}\) and carried a dish antenna with radius \(R=1.73 \mathrm{~m}\). The spacecraft was observed to decelerate with \(a=8 \times 10^{-10} \mathrm{~m} / \mathrm{s}^2\). It is known that some gaseous nebulae have densities of about \(10^{-16} \mathrm{~kg} / \mathrm{m}^3\). Was the Pioneer deceleration consistent with this number?

Short Answer

Expert verified
Answer: No, the observed deceleration is not consistent with the known density of gaseous nebulae, as the calculated deceleration due to the drag force is much smaller than the observed deceleration.

Step by step solution

01

Calculate the cross-sectional area

First, we need to calculate the cross-sectional area of Pioneer 10's dish antenna, which is a circle with radius \(R\). The formula for the area of a circle is \(A = \pi R^2\). Using \(R = 1.73 \mathrm{~m}\): $$A = \pi (1.73 \mathrm{~m})^2 \approx 9.41\mathrm{~m}^2$$
02

Convert the velocity

In order to have consistent units, we need to convert the velocity of Pioneer 10 from \(\mathrm{km/s}\) to \(\mathrm{m/s}\). Since \(1 \mathrm{~km} = 1000 \mathrm{~m}\), we have: $$v = 12.2 \mathrm{~km/s} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} = 12200 \mathrm{~m/s}$$
03

Calculate the drag force

Now we will calculate the drag force acting on Pioneer 10 using the formula \(F_d = \frac{1}{2} \rho v^2 A C_d\). We will assume that the drag coefficient \(C_d\) of the spacecraft is \(1\), as it is expected to be of order unity. Using the given density of gaseous nebulae \(\rho = 10^{-16} \mathrm{~kg/m^3}\), the velocity \(v=12200 \mathrm{~m/s}\), and the calculated area \(A \approx 9.41\mathrm{~m}^2\): $$F_d \approx \frac{1}{2}(10^{-16}\mathrm{~kg/m^3})(12200 \mathrm{~m/s})^2(9.41 \mathrm{~m^2})(1) \approx 6.83\times 10^{-9} \mathrm{~N}$$
04

Calculate the deceleration due to drag force

Now we will use Newton's second law \(F_d = ma\) to calculate the deceleration due to the drag force. The mass of Pioneer 10 is given as \(m = 240 \mathrm{~kg}\). Rearranging the formula and substituting the values: $$a = \frac{F_d}{m} \approx \frac{6.83\times 10^{-9} \mathrm{~N}}{240 \mathrm{~kg}} \approx 2.85\times 10^{-11}\mathrm{~m/s^2} $$
05

Compare the calculated deceleration with the observed deceleration

Finally, we will compare the calculated deceleration due to the drag force with the observed deceleration of \(a_{observed} = 8\times 10^{-10}\mathrm{~m/s^2}\). The calculated deceleration, \(2.85\times 10^{-11}\mathrm{~m/s^2}\), is much smaller than the observed deceleration, \(8\times 10^{-10}\mathrm{~m/s^2}\). This means that the observed deceleration of Pioneer 10 is not consistent with the known density of gaseous nebulae (\(10^{-16}\mathrm{~kg/m^3}\)). There might be other factors contributing to the deceleration of the spacecraft, which were not considered in this exercise.

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Most popular questions from this chapter

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