Question

How much work is required to lift a 1400-kg satellite to an altitude of 2⋅106 m above the surface of the Earth? The gravitational force is F=GMm/r2, where M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the satellite and the Earth's center. The radius of the Earth is 6.4⋅106 m, its mass is 6⋅1024 kg, and in these units the gravitational constant, G, is 6.67⋅10−11.

Answer 1

Answer:

2.08 x 10^10 J.

Step-by-step explanation:

Given:

m = 1400 kg

H = 2 x 10^6 m

R = 6.4 x 10^6 m

M = 6 x 10^24 kg

G = 6.67 x 10^-11

F = GMm/r^2

= GMm/r²

Work needed to lift a mass m from R (the surface) to R+h (height h above the surface) is just the difference of potential energy between those two altitudes. And potential energy is the r-integral of the force:

V(r) = -GMm/r

So you can calculate the lifting work from that.

Work, W = V(R+h) - V(R)

= GMm[1/R - 1/(R+h)]

Inputting values,

W = (6.67 × 10^−11) * (6 × 10^24) * (1400) * [(1/6.4×10^6) - (1/(6.4 × 10^6)+(2 × 10^6))]

= 2.08 x 10^10

W = 2.08 x 10^10 J.

Answer 2

The work done is required to lift a 1400-kg satellite to an altitude of 2⋅106 m  is mathematically given as

W = 2.08 x 10^10 J.

What is work is required to lift a 1400-kg satellite to an altitude of 2⋅106 m?

Question Parameters:

work is required to lift a 1400-kg satellite to an altitude of 2⋅106 m

The radius of the Earth is 6.4⋅106 m, its mass is 6⋅1024 kg,

the gravitational constant, G, is 6.67⋅10−11.

Generally, the equation for the   gravitational Force is mathematically given as

F = GMm/r^2

Where

V(r) = -GMm/r

Thereofore

W= GMm[1/R - 1/(R+h)]

W = (6.67 * 10^−11) * (6 * 10^24) * (1400) * [(1/6.4×10^6) - (1/(6.4 * 10^6)+(2 × 10^6)

W = 2.08 x 10^10 J.

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