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Chapter 8: Problem 3

When you stand on Earth, the distance between you and Earth is zero. So why isn't the gravitational force infinite?

Short Answer

Expert verified
The gravitational force isn't infinite because when calculating the force of gravity, the distance isn't the distance from the Earth's surface to an object but rather the distance from the Earth's center to the object.

Step by step solution

01

Understanding the Law of Universal Gravitation

Gravitational force is governed by the universally applicable law known as Newton's Law of Universal Gravitation, expressed as: \( F = G \frac{{m1 \cdot m2}}{{r^2}} \) where \(F\) is the force of attraction between the two bodies, \(m1\) and \(m2\) are the masses of the two bodies, \(r\) is the distance between the centers of the two bodies, and \(G\) is the gravitational constant.
02

Definition of 'Distance' in This Context

The 'distance' in the gravitational force formula considers the distance between the centers of the two masses, not the distance from the surface. Consequently, even if you're standing on the surface of the earth, you're still approximately 6,371 kilometers from its center.
03

Force Calculation

Therefore, when calculating the force of gravity acting on a body at the surface of the Earth, we do not take the distance as zero but rather the radius of the earth, which avoids the result of an 'infinite' force.

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Most popular questions from this chapter

Determine escape speeds from (a) Jupiter's moon Callisto and (b) a neutron star, with the Sun's mass crammed into a sphere of radius \(5.7 \mathrm{~km}\). See Appendix E for relevant data.

As a member of the 2040 Olympic committee, you're considering a new sport: asteroid jumping. On Earth, world-class high jumpers routinely clear \(2 \mathrm{~m}\). Your job is to make sure athletes jumping from asteroids will return to the asteroid. Make the simplifying assumption that asteroids are spherical, with average density \(1700 \mathrm{~kg} / \mathrm{m}^{3}\). For safety, make sure even a jumper capable of \(2.8 \mathrm{~m}\) on Earth will return to the surface. What do you report for the minimum asteroid diameter?

What's the approximate value of the gravitational force between a \(69-\mathrm{kg}\) astronaut and a \(77,000-\mathrm{kg}\) spacecraft when they're \(80 \mathrm{~m}\) apart?

Though the Sun's gravitational pull is much stronger than that of the Moon, the ocean tides on the Earth are caused principally by the Moon. Why?

Exact solutions for gravitational problems involving more than two bodies are notoriously difficult. One solvable problem involves a configuration of three equal-mass objects spaced in an equilateral triangle. Forces due to their mutual gravitation cause the configuration to rotate. Suppose three identical stars, each of mass \(M\), form a triangle of side \(L\) Find an expression for the period of their orbital motion.
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