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Chapter 4: Problem 49

At what altitude above the Earth's surface would your weight be half of what it is at the Earth's surface?

Short Answer

Expert verified
Explain your answer. Answer: No, it is not possible for someone to have half their weight at any altitude above Earth's surface. The solution resulted in a negative altitude, which indicates that the gravitational force does not decrease fast enough with increasing altitude to achieve half the weight before reaching the Earth's surface.

Step by step solution

01

Choose a weight for the person at Earth's surface

Let's label a person's weight (gravitational force) on the Earth's surface as \(F_1\) and the weight at the desired altitude as \(F_2\). We know that: \(F_2 = 0.5 F_1\) Let's choose a weight for the person on the Earth's surface. This doesn't matter, as it will cancel out in the final calculations. For example, let's say the person's weight is \(80 N\) on Earth's surface (\(F_1 = 80 N\)), and thus the weight (\(F_2\)) at the desired altitude should be \(40 N\).
02

Write down the equations for the weights at Earth's surface and desired altitude

Now we have the following equations for the gravitational force at Earth's surface and the desired altitude: \(F_1 = \cfrac{GMm}{R^2}\) \(F_2 = \cfrac{GMm}{(R+h)^2}\)
03

Find the relationship between the two equations

Divide the equation for \(F_2\) by the equation for \(F_1\) to find the relationship between the two equations: \(\cfrac{F_2}{F_1} = \cfrac{\cfrac{GMm}{(R+h)^2}}{\cfrac{GMm}{R^2}}\) We already know that \(F_2 = 0.5 F_1\), so: \(0.5 = \cfrac{(R+h)^2}{R^2}\)
04

Solve for the altitude (h)

Now, we need to solve for h in the equation: \(0.5 = \cfrac{(R+h)^2}{R^2}\) First, multiply both sides by \(R^2\): \(0.5R^2 = (R+h)^2\) Find the square root of both sides: \(\sqrt{0.5}R = R + h\) Now, solve for h: \(h = (\sqrt{0.5} - 1)R\) Plug in the Earth's radius (R) value: \(h = (6.371\times10^6m)(\sqrt{0.5} - 1)\) Calculate the altitude: \(h \approx -2.26\times10^6m\) This is a negative result, which means that it is not possible to have half of the weight at any altitude above the Earth's surface. This is because the gravitational force decreases with increasing altitude, but not at a fast enough rate to achieve half the weight before reaching the Earth's surface.

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Most popular questions from this chapter

In Fig. 4.15 an astronaut is playing shuffleboard on Earth. The puck has a mass of \(2.0 \mathrm{kg} .\) Between the board and puck the coefficient of static friction is 0.35 and of kinetic friction is \(0.25 .\) (a) If she pushes the puck with a force of \(5.0 \mathrm{N}\) in the forward direction, does the puck move? (b) As she is pushing, she trips and the force in the forward direction suddenly becomes \(7.5 \mathrm{N} .\) Does the puck move? (c) If so, what is the acceleration of the puck along the board if she maintains contact between puck and stick as she regains her footing while pushing steadily with a force of \(6.0 \mathrm{N}\) on the puck? (d) She carries her game to the Moon and again pushes a moving puck with a force of \(6.0 \mathrm{N}\) forward. Is the acceleration of the puck during contact more, the same, or less than on Earth? Explain. (tutorial: rough table)
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