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
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.

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Answer

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

Part (c): Velocity for circular orbit at radius RR

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

GMmR2=mv2R\frac{G M m}{R^{2}} = m \frac{v^{2}}{R}

Step 2: Cancel mm from both sides.

GMR2=v2R\frac{G M}{R^{2}} = \frac{v^{2}}{R}

Step 3: Multiply both sides by RR.

GMR=v2\frac{G M}{R} = v^{2}

Step 4: Take square root.

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

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

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

Step 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}

Step 2: Divide both sides by mm.

12v02−GMRe=−GMR\frac{1}{2} v_0^{2} - \frac{G M}{R_e} = -\frac{G M}{R}

Step 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}

Step 4: Factor out GMG M.

12v02=GM(1Re−1R)\frac{1}{2} v_0^{2} = G M \left( \frac{1}{R_e} - \frac{1}{R} \right)

Step 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)

Step 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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Part (c): Velocity for circular orbit at radius R Step 1: Gravitational force provides the centripetal force required for circular motion.

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

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI Answer
Part (c): Velocity for circular orbit at radius R Step 1: Gravitational force provides the centripetal force required for circular motion. (G M m)/(R^2) = m v^2R Step 2: Cancel m from both sides. (G M)/(R^2) = v^2R Step 3: Multiply both sides by R. (G M)/(R) = v^2 Step 4: Take square root. v = sqrt((G M)/(R)) Final answer for (c): sqrt((G M)/(R)) Part (d): Initial vertical velocity from Earth's surface (radius R_e) to reach distance R from center Step 1: Use conservation of mechanical energy. Kinetic + potential energy at surface equals potential energy at maximum distance R (where velocity is zero). (1)/(2) m v_0^2 - (G M m)/(R_e) = -(G M m)/(R) Step 2: Divide both sides by m. (1)/(2) v_0^2 - (G M)/(R_e) = -(G M)/(R) Step 3: Add (G M)/(R_e) to both sides. (1)/(2) v_0^2 = (G M)/(R_e) - (G M)/(R) Step 4: Factor out G M. (1)/(2) v_0^2 = G M ( (1)/(R_e) - (1)/(R) ) Step 5: Multiply both sides by 2. v_0^2 = 2 G M ( (1)/(R_e) - (1)/(R) ) Step 6: Take square root. v_0 = sqrt(2 G M ( (1)/(R_e) - (1)/(R) )) Final answer for (d): sqrt(2 G M ( (1)/(R_e) - (1)/(R) ))