Question

Scientists want to place a 4200 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2484 m/s in a perfectly circular orbit. Here is some information that may help solve this problem: mmars -6.4191 x 1023 kg mars -3.397 x 106 m G-6.67428 x 1011 N-m2/kg? m Submit 2) What is the force of attraction between Mars and the satellite? N Submit 3) What is the acceleration of the satellite in orbit? m/s Submit 4) Which of the following quantities would change the radius the satellite needs to orbit at? the mass of the satellite the mass of the planet Submit Othe speed of the satellite

Answer 1

Given is:-

Mass of the satellite

Orbital speed of the satellite  

mass of the mars   $m_m = 6.4191 \times 10^{23}kg$

radius of the mars   $r_m = 3.397 \times 10^6 m$

Now,

part - 1

The radius of the orbit is given by

$\frac{m_sv^2}{r} = \frac{Gm_sm_m}{r^2}$

by plugging all the values we get

$\frac{(4200 kg)(2484 m/s)^2}{r} = \frac{(6.67 \times 10^{-11}N-m^2/kg^2 )(4200 kg)(6.4191 \times 10^{23}kg)}{r^2}$

which gives us

$\boxed{r = 6.94 \times 10^6 m}$

Part -2 the force of attraction between the mars and the satellite is given by

$F = \frac{(6.67 \times 10^{-11}N-m^2/kg^2 )(4200 kg)(6.4191 \times 10^{23}kg)}{(6.94 \times 10^6m)^2}$

which gives us

$\boxed{F = 3733.62 N}$

Part - 3

The acceleration of the satellite is given by

$a = \frac{F}{m}$

thus

$a = \frac{3733.62N}{4200kg}$

which gives us

$\boxed{a =0.89 m/s^2}$

Part - 4

as we know that

$\frac{m_s v^2}{r} = \frac{Gm_sm_m}{r^2}$

or

$r = \frac{Gm_m}{v^2}$

hence,

the radius of the orbit would depend on

$\\\boxed{ \text{ mass of the planet}} \\ \boxed{\text{ speed of the satellite}}$

Part - 5

Time to take one revolution

$T = \frac{2 \pi r}{v}$

Now if we want the satelitte to take 8 time longer time, then

thus

$8 ( \frac{2 \pi r}{v}) = \frac{2 \pi r}{v'}$

which gives us

$v' = \frac{v}{8}$

by plugging the values of velcoity v of satellite we get

$v' = \frac{2484 m/s}{8}$

which gives us

$\boxed{v' =310.5 m/s}$