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2 parts
a

65.33 N65.33 \text{ N}

b

26384.80 s26384.80 \text{ s}

Step-by-step solution

8 steps

Here is the solution to the physics problem.

Given data:

  • Satellite mass, m=60 kgm = 60 \text{ kg}
  • Orbital radius, r=3Rr = 3R, where RR is the mean radius of the Earth
  • Gravitational field strength at Earth's surface, g=9.8 N/kgg = 9.8 \text{ N/kg}
  • Mean radius of the Earth, R=6400 km=6.4×106 mR = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}

Part a· 3 steps

Calculate the centripetal force acting on the satellite.

The centripetal force required to keep the satellite in orbit is provided by the gravitational force between the Earth and the satellite.
The gravitational force FgF_g is given by Newton's Law of Universal Gravitation:

Fg=GMmr2F_g = \frac{G M m}{r^2}

where GG is the gravitational constant and MM is the mass of the Earth.

The gravitational field strength at the Earth's surface is g=GMR2g = \frac{G M}{R^2}. From this, we can deduce that GM=gR2G M = g R^2.

  1. Substitute $G M$ into the gravitational force equation.

    The centripetal force FcF_c is equal to the gravitational force FgF_g:

    Fc=(gR2)mr2F_c = \frac{(g R^2) m}{r^2}

  2. Substitute the orbital radius r=3Rr = 3R.

    Since r=3Rr = 3R, then r2=(3R)2=9R2r^2 = (3R)^2 = 9R^2.

    Fc=gR2m9R2F_c = \frac{g R^2 m}{9R^2}

  3. Simplify the expression and calculate the centripetal force.

    The R2R^2 terms cancel out:

    Fc=gm9F_c = \frac{g m}{9} Fc=9.8 N/kg×60 kg9F_c = \frac{9.8 \text{ N/kg} \times 60 \text{ kg}}{9} Fc=5889 NF_c = \frac{588}{9} \text{ N} Fc=65.33 NF_c = \boxed{65.33 \text{ N}}

Part b

26384.80 s26384.80 \text{ s}

Calculate the period of the satellite in orbit.

The centripetal force can also be expressed in terms of the satellite's mass mm, its orbital speed vv, and the orbital radius rr:

Fc=mv2rF_c = \frac{m v^2}{r}

The orbital speed vv is related to the period TT by the relation v=2πrTv = \frac{2 \pi r}{T}.

  1. Substitute the expression for vv into the centripetal force equation.

    Fc=m(2πrT)2r=m(4π2r2/T2)r=4π2mrT2F_c = \frac{m \left(\frac{2 \pi r}{T}\right)^2}{r} = \frac{m (4 \pi^2 r^2 / T^2)}{r} = \frac{4 \pi^2 m r}{T^2}

  2. Equate the two expressions for centripetal force (Fc=gm9F_c = \frac{g m}{9} and Fc=4π2mrT2F_c = \frac{4 \pi^2 m r}{T^2}).

    4π2mrT2=gm9\frac{4 \pi^2 m r}{T^2} = \frac{g m}{9}

    We can cancel mm from both sides:

    4π2rT2=g9\frac{4 \pi^2 r}{T^2} = \frac{g}{9}

  3. Solve the equation for T2T^2.

    T2=4π2r×9g=36π2rgT^2 = \frac{4 \pi^2 r \times 9}{g} = \frac{36 \pi^2 r}{g}

  4. Substitute r=3Rr = 3R into the expression for T2T^2.

    T2=36π2(3R)g=108π2RgT^2 = \frac{36 \pi^2 (3R)}{g} = \frac{108 \pi^2 R}{g}

  5. Calculate the period TT.

    T=108π2Rg=π108RgT = \sqrt{\frac{108 \pi^2 R}{g}} = \pi \sqrt{\frac{108 R}{g}}

    Using R=6.4×106 mR = 6.4 \times 10^6 \text{ m} and g=9.8 m/s2g = 9.8 \text{ m/s}^2:

    T=π108×6.4×106 m9.8 m/s2T = \pi \sqrt{\frac{108 \times 6.4 \times 10^6 \text{ m}}{9.8 \text{ m/s}^2}} T=π691.2×1069.8T = \pi \sqrt{\frac{691.2 \times 10^6}{9.8}} T=π70530612.24T = \pi \sqrt{70530612.24} T≈3.14159×8398.25T \approx 3.14159 \times 8398.25 T≈26384.80 sT \approx \boxed{26384.80 \text{ s}}
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