Neglecting air resistance and the weight of the propellant, determine the work done in propelling a five-ton satellite to a height of (a) 100 miles above Earth and (b) 300 miles above Earth.

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)

= (8 × 10⁷)

= (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)

= (8 × 10⁷)

= (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

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2