Answer:
The value is 
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is 
The mass of the rover and the parachute is 
The drag coefficient is 
The atmospheric density of Earth is 
The acceleration due to gravity in Mars is 
Generally the Mars atmosphere density is mathematically represented as
=> 
=> 
Generally the drag force on the rover and the parachute is mathematically represented as
=> 
=> 
Gnerally this drag force is mathematically represented as
Here A is the frontal area
So
=> 
=>
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
Hello there.
<span>If we increase the force applied to an object and all other factors remain the same that amount of work will
</span><span>C. Increase
</span>
Answer:
C
Explanation:
Gravity is the main reason that make our planets to pull each other
Calm, sunny days with wind moving away from the center.
