Geostationary satellites are placed on orbits of radius 4.2×10^4m use this information to deduce g, the acceleration of freely falling body, at this height.
Ac = g there = v^2/R = omega^2 R
omega = 2 pi radians/24 hours
=2 pi radians /(24*3600)seconds
R is given
so find omega^2 R
Need solution
Answer
Physics
Well, if you're falling from a geostationary satellite, I hope you have a parachute made of helium-filled joke balloons! Because trust me, hitting the Earth from that height wouldn't be a fun way to end your journey.
But let's do some math humor here. To find the acceleration due to gravity (g) at that height, we can use the formula:
g = GM/R^2,
where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius at which you're located.
Now, using this information, we can substitute the given values and calculate the acceleration. Just make sure you hold onto something tightly, as the result might be shocking!
Calculating... "Eureka!"
*gasp*
Sorry, I got a little carried away there. It seems that the acceleration due to gravity at a height of 4.2×10^4m is approximately 6.67×10^-9 m/s^2. So, while falling from that height wouldn't be pleasant, at least you'd have a long time to make a few bad gravity puns on your way down!
To deduce the acceleration of a freely falling body at a height of a geostationary satellite, we can use the formula for gravitational acceleration:
g = G * M / r^2
Where:
g is the acceleration due to gravity,
G is the gravitational constant (approximately 6.674 x 10^-11 N*m^2/kg^2),
M is the mass of the Earth (approximately 5.97 x 10^24 kg),
r is the radius or distance from the center of the Earth.
In this case, the geostationary satellite is placed on an orbit of radius 4.2 × 10^4 m.
Plugging the values into the formula, we have:
g = (6.674 x 10^-11 N*m^2/kg^2) * (5.97 x 10^24 kg) / (4.2 × 10^4 m)^2
Simplifying this equation, we get:
g = 3.98 m/s^2
Therefore, the acceleration of a freely falling body at the height of a geostationary satellite is approximately 3.98 m/s^2.