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Chapter 14: Problem 3

One of the Echo satellites consisted of an inflated spherical aluminum balloon \(30 \mathrm{~m}\) in diameter and of mass \(20 \mathrm{~kg}\). Suppose a meteor having a mass of \(7.0 \mathrm{~kg}\) passes within \(3.0 \mathrm{~m}\) of the surface of the satellite. What is the magnitude of the gravita- tional force on the meteor from the satellite at the closest approach?

Short Answer

Expert verified
The gravitational force on the meteor is approximately \(2.88 \times 10^{-12} \mathrm{N}\).

Step by step solution

01

- Identify the masses and distance

Identify the given values: Mass of satellite, \(M = 20 \mathrm{kg}\); Mass of meteor, \(m = 7.0 \mathrm{kg}\); Distance of closest approach, which is the radius plus the additional distance \(d = \frac{30 \mathrm{m}}{2} + 3.0 \mathrm{m} = 18 \mathrm{m}\).
02

- Use the universal law of gravitation

The gravitational force between two masses is given by Newton's law of gravitation: \[ F = G \frac{Mm}{r^2} \] where \( G = 6.674 \times 10^{-11} \mathrm{N \, (m/kg)^2} \).
03

- Plug in the values

Substitute the given values into the formula: \[ F = 6.674 \times 10^{-11} \frac{(20)(7.0)}{18^2} \]
04

- Calculate the force

Compute the value: \[ F = 6.674 \times 10^{-11} \frac{140}{324} \approx 2.88 \times 10^{-12} \mathrm{N} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's law of gravitation
Newton's law of gravitation is fundamental for understanding how masses interact. It states that every mass exerts a force on every other mass. This force is attractive and acts along the line connecting the two masses. The formula is:









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    M and m are the masses involved.


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    This law explains gravity's role in our universe, from falling apples to orbiting planets. Understanding it is key to solving questions about gravitational forces.
Mass and distance
The magnitude of the gravitational force depends heavily on the mass of the objects and the distance between them.

Here鈥檚 why:

    This means heavier objects exert a stronger gravitational pull.







        Often, we need to add extra distance to account for the shapes or sizes of objects involved. Like in the given exercise, the meteor's closest distance took into account the satellite's radius plus additional distance. Understanding this helps us calculate gravity correctly.
Universal gravitational constant
The universal gravitational constant, denoted as G, is a key part of Newton's law of gravitation. Its value is approximately

Let's break down its significance:

    *Bounce between planets*
    Understanding G allows modern science to predict how objects as vast as planets and stars interact. It's vital for fields like astrophysics and engineering. In the given exercise, using the correct G value ensures your calculations are accurate. Getting a solid grasp on it, boosts your ability to tackle similar problems confidently.

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Most popular questions from this chapter

In deep space, sphere \(A\) of mass \(20 \mathrm{~kg}\) is located at the origin of an \(x\) axis and sphere \(B\) of mass \(10 \mathrm{~kg}\) is located on the axis at \(x=0.80 \mathrm{~m} .\) Sphere \(B\) is released from rest while sphere \(A\) is held at the origin. (a) What is the gravitational potential energy of the two-sphere system as \(B\) is released? (b) What is the kinetic energy of \(B\) when it has moved \(0.20 \mathrm{~m}\) toward \(A\) ?

Is Newton's Law of Gravity Wrong? A professional scientist (not a physicist) stops you in the hall and says: "I can prove Newton's theory of gravity is wrong. The Sun is 320,000 times as massive as the Earth, but only 400 times as far from the Moon as is the Earth. Therefore, the force of the Sun's gravity on the Moon should be twice as big as the Earth's and the Moon should go around the Sun instead of around the Earth. Since it doesn't, Newton's theory of gravity must be wrong!" What's the matter with this reasoning?

A solid sphere of uniform density has a mass of \(1.0 \times 10^{4} \mathrm{~kg}\) and a radius of \(1.0 \mathrm{~m}\). What is the magnitude of the gravitational force due to the sphere on a particle of mass \(m\) located at a distance of (a) \(1.5 \mathrm{~m}\) and (b) \(0.50 \mathrm{~m}\) from the center of the sphere? (c) Write a general expression for the magnitude of the gravitational force on the particle at a distance \(r \leq 1.0 \mathrm{~m}\) from the center of the sphere.

One model for a certain planet has a core of radius \(R\) and mass \(M\) surrounded by an outer shell of inner radius \(R\), outer radius \(2 R\), and mass \(4 M\). If \(M=4.1 \times 10^{24} \mathrm{~kg}\) and \(R=6.0 \times 10^{6} \mathrm{~m}\), what is the gravitational acceleration of a particle at points (a) \(R\) and (b) \(3 R\) from the center of the planet?

Rate of Rotation The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. (Why?) (a) Show that the corresponding shortest period of rotation is $$ T=\sqrt{\frac{3 \pi}{G \rho}} $$ where \(\rho\) is the uniform density of the spherical planet. (b) Calculate the rotation period assuming a density of \(3.0 \mathrm{~g} / \mathrm{cm}^{3}\), typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis.
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