Question

Putting a Satellite into Orbit The strength of Earth's gravitational field varies with the distance \(r\) from Earth's center, and the magnitude of the gravitational force experienced by a satellite of mass \(m\) during and after launch is $$F(r)=\frac{m M G}{r^{2}}$$ Here, \(M=5.975 \times 10^{24} \mathrm{kg}\) is Earth's mass, \(G=6.6726 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} \mathrm{kg}^{-2}\) is the universal gravitational constant, and \(r\) is measured in meters. The work it takes to lift a 1000 -kg satellite from Earth's surface to a circular orbit \(35,780 \mathrm{km}\) above Earth's center is therefore given by the integral Work $$=\int_{6,370,000}^{35.780,000} \frac{1000 M G}{r^{2}} d r$$ joules. The lower limit of integration is Earth's radius in meters at the launch site. Evaluate the integral. (This calculation does not take into account energy spent lifting the launch vehicle or energy spent bringing the satellite to orbit velocity.)

Short answer

Do the final calculation to find out the work done to lift the satellite to the specified height in joules.

Step-by-step solution

Understanding the Integral

Since the work done to lift a satellite is equivalent to the definite integral of the force over the distance lifted, we need to compute the integral of \(F(r) =\frac{1000 M G}{r^{2}}\) from \(r = 6,370,000\) to \(r = 35,780,000\).

Substituting Known Quantities

Replace the constants \(M = 5.975 \times 10^{24} \mathrm{kg}\) and \(G = 6.6726 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} \mathrm{kg}^{-2}\) into the equation of the force. This will yield: \(F(r) = \frac{1000 \times 5.975 \times 10^{24} \times 6.6726 \times 10^{-11}}{r^{2}}\) .

Integral Calculation

Now, find the integral: \( \int_{6,370,000}^{35,780,000} F(r) dr \). The integral of \(1/r^{2}\) over \(r\) is \(-1/r\). Therefore, the solution of the integral is: \(- \frac{1000 \times 5.975 \times 10^{24} \times 6.6726 \times 10^{-11}}{r} \biggr|_{6,370,000}^{35,780,000}\).

Evaluating the Integral at Limits

Substitute the upper and lower limits into the resulting expression. This yields: \(- \frac{1000 \times 5.975 \times 10^{24} \times 6.6726 \times 10^{-11}}{35,780,000} + \frac{1000 \times 5.975 \times 10^{24} \times 6.6726 \times 10^{-11}}{6,370,000}\).

Final Calculation

Finally, calculate the result of the above expression. The obtained result will represent the work done to put the satellite into orbit in terms of joules.