Question 12.SP.14: A satellite is launched in a direction parallel to the earth......

STRATEGY: After the satellite is launched, it is subjected to the earth’s gravitational attraction only and undergoes central-force motion. Knowing this, you can determine the satellite’s trajectory, maximum altitude, and periodic time.

MODELING and ANALYSIS: The satellite can be modeled as a particle.

a. Maximum Altitude. After the satellite is launched, it is subject only to the earth’s gravitational attraction. Thus, its motion is governed by Eq. (12.37), so

{\frac{1}{r}}=u={\frac{G M}{h^{2}}}+C\cos\theta           (12.37)

{\frac{1}{r}}={\frac{G M}{h^{2}}}+C\cos\theta           (1)

Because the radial component of the velocity is zero at the point of launching A, you have h =r_{0}v_{0}. Recalling that for the earth, R = 6370 km, you can compute

r_{\mathrm{0}} = 6370 km + 500 km = 6870 km = 6.87 × 10^{6}\, m

v_{\mathrm{0}} = 36 900 km/h = \frac{36.9\times10^{6}\mathrm{{~m}}}{3.6\times10^{3}\mathrm{{~s}}}= 10.25 × 10³ m/s

h=r_{0}v_{0}=(6.87\times10^{6}\,{\mathrm{m}})(10.25\times10^{3}\,{\mathrm{m}}/{\mathrm{s}})=70.4\times10^{9}\,{\mathrm{m}}^{2}/{\mathrm{s}}

h^{2}=4.96\times10^{21}\,\mathrm{m}^{4}/\mathrm{s}^{2}

Because GM = gR², where R is the radius of the earth, you also have

G M=g R^{2}=(9.81\;\mathrm{m/s}^{2})(6.37\times10^{6}\,\mathrm{m})^{2}=398\times10^{12}\,\mathrm{m^{3}/s^{2}}

{\frac{G M}{h^{2}}}={\frac{398\times10^{12}\,{\mathrm{m}}^{3}/s^{2}}{4.96\times10^{21}\,{\mathrm{m}}^{4}/{\mathrm{s}}^{2}}}=80.3\times10^{-9}\,{\mathrm{m}}^{-1}

Substituting this value into Eq. (1) gives

{\frac{1}{r}}=80.3\times10^{-9}\,\mathrm{m}^{-1}+C\cos\theta               (2)

Note that at point A, θ = 0 and r = r_{0} = 6.87 × 10^{6}m (Fig. 1). From this, you can compute the constant C as

\frac{1}{6.87\times10^{6}\,{\mathrm{m}}}=80.3\times10^{-9}\,{\mathrm{m}}^{-1}+C\,\mathrm{cos~0^{\circ}}\qquad C=65.3\times10^{-9}\,{\mathrm{m}}^{-1}

At A′, which is the point on the orbit farthest from the earth, you have θ = 180° (Fig. 1). Using Eq. (2), you can compute the corresponding distance r_{1} to be

{\frac{1}{r_{1}}}=80.3\times10^{-9}\,\mathrm{m}^{-1}+(65.3\times10^{-9}\,\mathrm{m}^{-1})\,\mathrm{cos}\,\,180^{\circ}

r_{1}=66.7\times10^{6}\,{\mathrm{m}}=66\,700\,{\mathrm{km}}

Maximum altitude = 66 700 km − 6370 km = 60 300 km ◂

b. Periodic Time. Because A and A′ are the perigee and apogee, respectively, of the elliptical orbit, use Eqs. (12.44) and (12.45) to compute the semimajor and semiminor axes of the orbit (Fig. 2):

a=\frac{1}{2}\left(r_0+r_1\right)            (12.44)

b=\sqrt{r_0 r_1}            (12.45)

a={\textstyle\frac{1}{2}}(r_{0}+r_{1})={\textstyle\frac{1}{2}}(6.87+66.7)(10^{6})\,\mathrm{m}=36.8\times10^{6}\,\mathrm{m}

b=\sqrt{r_{0}r_{1}}=\sqrt{(6.87)(66.7)}\times10^{6}\,{\mathrm{m}}=21.4\times10^{6}\,{\mathrm{m}}

\tau={\frac{2\pi a b}{h}}={\frac{2\pi(36.8\times10^{6}\operatorname{m})(21.4\times10^{6}\operatorname{m})}{70.4\times10^{9}\operatorname{m}^{2}/s}}

τ = 70.3 × 10³s = 1171 min = 19 h 31 min ◂

REFLECT and THINK: The satellite takes less than one day to travel over 60 000 km around the earth. In this problem, you started with Eq. 12.37, but it is important to remember that this formula was the solution to a differential equation that was derived using Newton’s second law.