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2 parts
a

31.392 N\text{31.392 N}

b

56797.8 s\text{56797.8 s}

Step-by-step solution

10 steps

Here's the information provided:

  • Mass of the satellite, m=80 kgm = 80 \text{ kg}
  • Orbital radius, r=5Rr = 5R, where RR is the mean radius of the Earth.
  • Gravitational field strength at the Earth's surface, g=9.81 N/kgg = 9.81 \text{ N/kg}
  • Radius of the Earth, R=6400 km=6.4×106 mR = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}

Part a· 4 steps

Calculate the centripetal force

  1. The centripetal force required to keep the satellite in orbit is provided by the gravitational force.

    The gravitational force is given by Newton's Law of Universal Gravitation:

    Fg=GMmr2F_g = \frac{GMm}{r^2}

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

  2. We know that the gravitational field strength at the Earth's surface is g=GMR2g = \frac{GM}{R^2}.

    From this, we can express GMGM as GM=gR2GM = gR^2.

  3. Substitute this expression for GMGM into the gravitational force formula.

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

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

    Given that the orbital radius is r=5Rr = 5R, substitute this value:

    Fc=gR2m(5R)2F_c = \frac{gR^2m}{(5R)^2} Fc=gR2m25R2F_c = \frac{gR^2m}{25R^2}

    The R2R^2 terms cancel out:

    Fc=gm25F_c = \frac{gm}{25}

  4. Now, substitute the numerical values

    Fc=9.81 N/kg×80 kg25F_c = \frac{9.81 \text{ N/kg} \times 80 \text{ kg}}{25} Fc=784.825F_c = \frac{784.8}{25} Fc=31.392 NF_c = 31.392 \text{ N}

    The centripetal force is 31.392 N\boxed{\text{31.392 N}}.

Part b

56797.8 s\text{56797.8 s}

Calculate the period of the satellite in orbit

  1. For a satellite in a circular orbit, the centripetal force is equal to the gravitational force.

    The centripetal force can also be expressed as Fc=mv2rF_c = \frac{mv^2}{r}, where vv is the orbital speed of the satellite.
    So, we have:

    mv2r=GMmr2\frac{mv^2}{r} = \frac{GMm}{r^2}

    Simplifying mm and rr:

    v2=GMrv^2 = \frac{GM}{r}

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

    Squaring this gives:

    v2=4π2r2T2v^2 = \frac{4\pi^2 r^2}{T^2}

  3. Equate the two expressions for v2v^2

    4π2r2T2=GMr\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}

    Rearrange to solve for T2T^2:

    T2=4π2r3GMT^2 = \frac{4\pi^2 r^3}{GM}

  4. Substitute GM=gR2GM = gR^2 and r=5Rr = 5R into the equation for T2T^2

    T2=4π2(5R)3gR2T^2 = \frac{4\pi^2 (5R)^3}{gR^2} T2=4π2(125R3)gR2T^2 = \frac{4\pi^2 (125R^3)}{gR^2} T2=500π2RgT^2 = \frac{500\pi^2 R}{g}

  5. Take the square root to find TT

    T=500π2RgT = \sqrt{\frac{500\pi^2 R}{g}} T=π500RgT = \pi \sqrt{\frac{500R}{g}}

  6. Now, substitute the numerical values

    T=π500×6.4×106 m9.81 N/kgT = \pi \sqrt{\frac{500 \times 6.4 \times 10^6 \text{ m}}{9.81 \text{ N/kg}}} T=π3200×1069.81T = \pi \sqrt{\frac{3200 \times 10^6}{9.81}} T=π326197757.39T = \pi \sqrt{326197757.39} T=π×18060.9456T = \pi \times 18060.9456 T≈56797.8 sT \approx 56797.8 \text{ s}

    The period of the satellite is approximately 56797.8 s\boxed{\text{56797.8 s}}.

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