Question

an astronomy textbook, when printed out, weighs four pounds on the surface of the earth. after finishing your course, you are so tired of the book, you arrange for nasa to shoot it into space. when it is twice as far from the center of the earth than when you were reading it, what would it weigh? (note: assume that the book has been moving away from the earth, not falling freely around it.) problem 3 16 lbs 8 lbs 4 lbs 2 lbs 1 lb hint feedback i give up! submission(s) remaining hints: 1 deduction per hint. hints remaining: 1 feedback: 1% deduction per feedback. grade summary deductions potential submissions attempt(s) remaining 1% deduction per detailed view

Explanation:

Step1: Recall gravitational - force formula

The gravitational force \(F = G\frac{Mm}{r^{2}}\), where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, \(m\) is the mass of the object, and \(r\) is the distance from the center of the Earth to the object. On the surface of the Earth, let \(F_1=mg_1 = G\frac{Mm}{R^{2}}\), where \(R\) is the radius of the Earth and \(g_1\) is the acceleration due to gravity on the Earth's surface.

Step2: Analyze the new - distance case

When the object is at a distance \(r = 2R\) (twice as far from the center of the Earth as on the surface), the gravitational force \(F_2=G\frac{Mm}{(2R)^{2}}=\frac{1}{4}G\frac{Mm}{R^{2}}\). Since \(F_1 = mg_1=G\frac{Mm}{R^{2}}\) and \(F_2 = mg_2\), then \(F_2=\frac{1}{4}F_1\).

Step3: Calculate the new weight

The weight of the book on the surface of the Earth is \(F_1 = 4\) lbs. The new weight \(F_2\) at a distance twice as far from the center of the Earth is \(\frac{1}{4}\times4 = 1\) lb.

Answer:

1 lb