Question

5. O planeta Marte possui massa de 6,46 . 1023 kg e raio 3,37 . 106m. Sengo G = 6,66 . 10-11 N.m²/kg² a constante de gravitação universal, determine:

a. A velocidade de escape nesse planeta;

b. A velocidade orbital e o período de um satélite artificial que orbite em baixa altitude (satélite rasante) nesse planeta (raio da óbita – raio de Marte).​

Answer 1

5)

Dados:

$M=6.46\times10^{23}\ kg\\ r=3.37\times10^{6}\ m\\ G\approx6.66\times10^{-11}\ Nm^2kg^{-2}$

a)

$\boxed{v_e=\sqrt{\dfrac{2GM}{r}}}\ \to\ v_e=\sqrt{\dfrac{2(6.66\times10^{-11})(6.46\times10^{23})}{3.37\times10^{6}}}\ \therefore$

$v_e=\sqrt{\dfrac{86.0472\times10^{12}}{3.37\times10^6}}\approx\sqrt{25.5333\times10^6}\ \therefore$

$\boxed{v_e\approx5.05305\times10^3\ m/s}$

b)

$\boxed{v_{orbital}=\sqrt{\dfrac{GM}{r}}\ \therefore\ v_{orbital}=\dfrac{v_e}{\sqrt{2}}}\ \to$

$v_{orbital}\approx\dfrac{5.05305\times10^3}{\sqrt{2}}\ \therefore\ \boxed{v_{orbital}\approx3.57305\times10^3\ m/s}$

Para um satélite rasante, teremos:

$F_G=F_{cp}\ \therefore\ \dfrac{GMm}{r^2}=m\omega^2r\ \therefore\ \omega^2=\dfrac{GM}{r^3}\ \therefore$

$\bigg(\dfrac{2\pi}{T}\bigg)^2=\dfrac{GM}{r^3}\ \therefore\ T^2=\dfrac{4\pi^2r^3}{GM}\ \therefore\ \boxed{T=2\pi r\sqrt{\dfrac{r}{GM}}}\ \therefore$

$\boxed{T=\dfrac{2\pi r}{v_{orbital}}}\ \therefore\ T\approx\dfrac{2\pi(3.37\times10^6)}{3.57305\times10^3}\ \therefore\ \boxed{T\approx5.92612319\times10^3\ s}$