A space capsule weighing 5000 pounds is propelled to an altitude of 200 miles above the surface of the earth. How much work is done against the force of gravity? Assume the earth is a sphere of radius 4000 miles and that the force of gravity is f(x)=-k/x^2, where x is the distance from the center of the earth to the capsule (the inverse-square law). Thus, the lifting force required is k/x^2, and this equals 5000 when x=4000. (Leave answer in units mile-pounds; do not convert to feet.)

Where W is the work done, F(x) is the force at a distance x from the center of the Earth, and r1 and r2 are the initial and final distances from the center of the Earth, respectively.

The force of gravity is given by F(x) = -k/x^2. We also know that the lifting force required to counteract the force of gravity is 5000 pounds when the distance x from the center of the Earth is 4000 miles.

We can calculate k using this information:

-5000 = -k/4000^2

-5000 = -k/16000000

k = 5000 * 16000000

k = 80000000000

Now we can find the work done from the initial height (r1 = 4000 miles) to the final height (r2 = 4000 + 200 miles).

W = ∫[4000 to 4200] (-80000000000/x^2) dx

W = -80000000000 ∫[4000 to 4200] (1/x^2) dx

W = -80000000000 * (-1/x)[4000 to 4200]

W = 80000000000 * (1/4000 - 1/4200)

W = 80000000000 * (0.00025 - 0.0002381)

W = 80000000000 * 0.0000119

W = 952000 mile-pounds

Therefore, the work done against the force of gravity is 952000 mile-pounds.

msm555 msm555 · Answer 2 · 2024-01-21T01:50:02+01:00

Answer:

952380.95 mile-pounds

Step-by-step explanation:

We are given a capsule weighing 5000 pounds, and our task is to lift it 200 miles above the surface of the earth. Assume the earth is a sphere whose radius is 4000 miles, and the gravitational force is given by [tex] f(x) = -\dfrac{k}{x^2} [/tex], where [tex] x [/tex] is the distance from the center of the earth.

We will use the opposite force [tex] f(x) [/tex] to find the work done in lifting the capsule.

To determine the value of [tex] k [/tex], we can use the given information that when [tex] x = 4000 [/tex],

[tex] f(4000) = -\dfrac{k}{4000^2} = 5000 [/tex].

Solving for [tex] k [/tex]:

[tex] -\dfrac{k}{4000^2} = 5000 [/tex]

[tex] k = -8 \times 10^{10} [/tex]

Now, the work done in lifting the capsule from an initial distance of 4000 miles to a final distance of 4200 miles can be calculated using the formula for work:

[tex] W = \big \int_{4000}^{4200} -\dfrac{k}{x^2} \,dx [/tex]

Substitute the value of [tex] k [/tex]:

[tex] W =- 8 \times 10^{10} \int_{4000}^{4200} -\dfrac{1}{x^2} \,dx [/tex]

Now, calculate the definite integral:

[tex] W = 8 \times 10^{10} \left[ \dfrac{-1}{x} \right]_{4000}^{4200} [/tex]

[tex] W = 8 \times 10^{10} \left( \dfrac{-1}{4200} - \dfrac{-1}{4000} \right) [/tex]

[tex] W \approx 952380.9524 \, \text{mile-pounds} [/tex]

[tex] W \approx 952380.95 \, \text{mile-pounds ( in nearest 2 d p.)} [/tex]

Therefore, the work done in lifting the capsule is approximately 952380.95 mile-pounds.