A given 15000 N on earth 2250 N on moon Moon Radius 1.74 x 10 6 m Orbital Height 100km Total...


A given 15000 N on earth 2250 N on moon Moon Radius= {eq}1.74x10^{6}m {/eq} Orbital Height 100km Total orbit {eq}1.84x10^{6}m {/eq}

What is orbital speed?

Stable Orbit

If a mass m is in a stable circular orbit, then the acceleration due gravity is the same as the necessary centripetal acceleration for that mass m to have the tangential velocity {eq}v_t {/eq} of the given orbital radius r.

Centripetal acceleration:

{eq}a_c=\dfrac{v_t^2}{r} {/eq}

Gravitational acceleration:

{eq}g=\frac{GM}{r^2} {/eq}

The gravitational constant is:

{eq}G=6.674\times 10^{-11} \frac{m^3}{kg \cdot s^2} {/eq}

Answer and Explanation:

Weight is mass times the gravitational acceleration.

{eq}F=mg {/eq}

So given the two weights of the object, we can determine the ratio between the gravitational acceleration on each body.

{eq}\frac{F_E}{F_m}=\frac{g_E}{g_m}=g_E\frac{R_m^2}{GM_m} {/eq}

We can rearrange this to solve for the mass of the moon.

{eq}M_m=\frac{g_ER_m^2}{G}\frac{F_m}{F_E}=\frac{(9.81)(1.74 \times 10^6)^2}{(6.674\times 10^{-11})}\frac{2250}{15000}=6.68 \times 10^{22} kg {/eq}

If a mass were in orbit around the moon with an orbital height of 100 km, then the orbital radius would be that height plus the radius of the moon. The gravitational acceleration at that radius would be:

{eq}g=\frac{GM_m}{(R_m+h)^2}=\frac{(6.674\times 10^{-11})(6.68 \times 10^{22})}{(1.74 \times 10^6+1 \times 10^5)^2}=1.32 \frac{m}{s^2} {/eq}

Setting this equal to the centripetal acceleration:

{eq}\frac{GM_m}{(R_m+h)^2}=\frac{v_t^2}{(Rm+h)} {/eq}

Solve for tangential velocity:

{eq}v_t=\sqrt{\frac{GM_m}{R_m+h}}=\sqrt{\frac{(6.674\times 10^{-11})(6.68 \times 10^{22})}{(1.74 \times 10^6+1 \times 10^5)}}=1560 \frac{m}{s} {/eq}

Learn more about this topic:

Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16

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