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An earth satellite describes a circular orbit of radius R about the center of the earth. Calculate R if the orbit period is one day. Calculate initial vertical velocity to leave the satellite from the surface to a distance R from the center. Calculate launching velocity for a circular orbit, assuming earth's rotation is not considered.

Physics

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Answers

2 parts
c

GMR\sqrt{\dfrac{G M}{R}}

d

2GM(1Re−1R)\sqrt{2 G M \left( \dfrac{1}{R_e} - \dfrac{1}{R} \right)}

Step-by-step solution

10 steps

Part c· 4 steps

Velocity for circular orbit at radius RR

  1. Gravitational force provides the centripetal force required for circular motion.

    GMmR2=mv2R\frac{G M m}{R^{2}} = m \frac{v^{2}}{R}
  2. Cancel mm from both sides.

    GMR2=v2R\frac{G M}{R^{2}} = \frac{v^{2}}{R}
  3. Multiply both sides by RR.

    GMR=v2\frac{G M}{R} = v^{2}
  4. Take square root.

    v=GMRv = \sqrt{\frac{G M}{R}}

    Final answer for (c): GMR\boxed{\sqrt{\dfrac{G M}{R}}}

Part d

2GM(1Re−1R)\sqrt{2 G M \left( \dfrac{1}{R_e} - \dfrac{1}{R} \right)}

Initial vertical velocity from Earth's surface (radius ReR_e) to reach distance RR from center

  1. Use conservation of mechanical energy.

    Kinetic + potential energy at surface equals potential energy at maximum distance RR (where velocity is zero).

    12mv02−GMmRe=−GMmR\frac{1}{2} m v_0^{2} - \frac{G M m}{R_e} = -\frac{G M m}{R}
  2. Divide both sides by mm.

    12v02−GMRe=−GMR\frac{1}{2} v_0^{2} - \frac{G M}{R_e} = -\frac{G M}{R}
  3. Add GMRe\frac{G M}{R_e} to both sides.

    12v02=GMRe−GMR\frac{1}{2} v_0^{2} = \frac{G M}{R_e} - \frac{G M}{R}
  4. Factor out $G M$.

    12v02=GM(1Re−1R)\frac{1}{2} v_0^{2} = G M \left( \frac{1}{R_e} - \frac{1}{R} \right)
  5. Multiply both sides by 2.

    v02=2GM(1Re−1R)v_0^{2} = 2 G M \left( \frac{1}{R_e} - \frac{1}{R} \right)
  6. Take square root.

    v0=2GM(1Re−1R)v_0 = \sqrt{2 G M \left( \frac{1}{R_e} - \frac{1}{R} \right)}

    Final answer for (d): 2GM(1Re−1R)\boxed{\sqrt{2 G M \left( \dfrac{1}{R_e} - \dfrac{1}{R} \right)}}

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