Answer:
a) the work done in propelling a five-ton satellite to a height of 100 miles above Earth is 487.8 mile-tons
b) the work done in propelling a five-ton satellite to a height of 300 miles above Earth is 1395.3 mile-tons
Step-by-step explanation:
Given the data in the question;
We know that the weight of a body varies inversely as the square of its distance from the center of the earth.
⇒F(x) = c / x²
given that; F(x) = five-ton = 5 tons
we know that the radius of earth is approximately 4000 miles
so we substitute
5 = c / (4000)²
c = 5 × ( 4000 )²
c = 8 × 10⁷
∴ Increment of work is;
Δw = [ ( 8 × 10⁷ ) / x² ] Δx
a) For 100 miles above Earth;
W = ₄₀₀₀∫⁴¹⁰⁰ [ ( 8 × 10⁷ ) / x² ] Δx
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{x}[/tex] [tex]]^{4100}_{4000[/tex]
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{4100}[/tex] [tex]+\frac{1}{4000}[/tex] [tex]][/tex]
= (8 × 10⁷ ) [ 6.09756 × 10⁻⁶ ]
= 487.8 mile-tons
Therefore, the work done in propelling a five-ton satellite to a height of 100 miles above Earth is 487.8 mile-tons
b) For 300 miles above Earth.
W = ₄₀₀₀∫⁴³⁰⁰ [ ( 8 × 10⁷ ) / x² ] Δx
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{x}[/tex] [tex]]^{4300}_{4000[/tex]
= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{4300}[/tex] [tex]+\frac{1}{4000}[/tex] [tex]][/tex]
= (8 × 10⁷ ) [ 1.744186 × 10⁻⁵ ]
= 1395.3 mile-tons
Therefore, the work done in propelling a five-ton satellite to a height of 300 miles above Earth is 1395.3 mile-tons
