Answer to Question #180217 in Classical Mechanics for Sami ullah babar

Question #180217

When the Mars Exploration Rover was fully assembled, its mass was 180 kg. The acceleration due 

to gravity at the surface of Mars is 3.68 m/s2and the radius of Mars is 3390 km. 

(a) What was the rover s weight when it was at sea level on Earth? 

(b) What is the rover s weight on the surface of Mars? 

(c) The entry phase began when the spacecraft reached the Mars atmospheric entry interface 

point at 3522 km from the center of Mars. What was the rover’s weight at that point?


1
Expert's answer
2021-04-13T06:31:52-0400

Given,

Mass of the rover = 180 kg

Acceleration due to gravity at the surface of mars "(a_m)=3.68 m\/s^2"

Radius of mars "(R)=3390 km."

Acceleration due to gravity at the surface of earth "(g_e)=9.8 m\/s^2"

a) Rover's weight at the surface of earth, "(W)= mg = 180\\times 9.8 N"

"=1764N"

b) Rover's weight at the surface of mars "(W_m)=ma_m =180\\times 3.68 N"

"=662.4N"

c)"g_m=\\frac{GM}{R^2}"

Gravitational field at any distance from the surface of mars

"g_m=\\frac{GM}{(R+h)^2}"

"\\Rightarrow g_m=\\frac{GM}{R^2(1+\\frac{h}{R})^2}"

Hence, rover's weight at the given distance,

"W'=mg_m"

Now, substituting the values,

"W'=180\\times \\frac{3.68}{(1+\\dfrac{3522-3390}{3390})^2}" N


"\\Rightarrow W'=\\frac{180\\times3.68}{(1+\\dfrac{132}{3390})^2}" N


"\\Rightarrow W'=\\frac{662.4N}{(1+0.0389)^2}"


"\\Rightarrow W' =\\frac{662.4N}{(1.079)}"


"\\Rightarrow W' =613.90N"


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