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Chapter 14: Problem 3

One of the Echo satellites consisted of an inflated aluminum balloon \(30 \mathrm{~m}\) in diameter and of mass \(20 \mathrm{~kg}\). A meteor having a mass of \(7.0 \mathrm{~kg}\) passes within \(3.0 \mathrm{~m}\) of the surface of the satellite. If the effect of all bodies other than the meteor and satellite are ignored, what gravitational force does the meteor experience at closest approach to the satellite?

Short Answer

Expert verified
The gravitational force that the meteor experiences at closest approach to the satellite is \(F = 1.04 \times 10^{-9} \: N\).

Step by step solution

01

Identify the known variables.

We know that the mass of the satellite (m1) is 20 kg, the mass of the meteor (m2) is 7.0 kg, the diameter of the satellite d is 30 m, therefore the radius (r1) is 15 m, and the distance of closest approach (d) is 3.0 m.
02

Calculate the total distance between centers of the meteor and the satellite (r).

Since the meteor passes with 3 m from the surface of the satellite, the total distance (r) between their centers is the sum of the satellite radius (r1) and the closest approach (d). So, \(r = r1 + d = 15m + 3m = 18m\).
03

Calculate the gravitational force (F).

Now we can substitute the known values into the universal gravitational force formula. The gravitational force (F) then is: \(F = G \cdot (m1 \cdot m2) / r^2 = 6.674 \times 10^{-11} \: N \cdot m^2/kg^2 \cdot (20kg \cdot 7kg) / (18m)^2\). After making the calculation we find that \(F = 1.04 \times 10^{-9} \: N\).

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Most popular questions from this chapter

Two point-like particles, each of mass \(m\), are originally separated by a distance \(d\) and moving in opposite directions each with a speed of \(v\). What is the maximum value for \(v\) so that the particles will eventually move back together under the influence of the mutual gravitational attraction?

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