Assuming Earth behaves as a perfect sphere with a radius of 6380 km, the standard gravitational acceleration,g, at sea level has a value of 9.80 m/s^2.
a.) As Earth rotates on its axis, objects at the equator experience a slightly different value of g due to additional centripetal acceleration. Calculate this adjusted value for g.
b.) Is this difference due to centripetal forces present at the poles? If g is dependent upon the position on the surface of Earth, why do we use the standard value of 9.80 m/s^2 in our calculations? Explain.
c.) Whats the value of g on board a space shuttle in a stable orbit at a height of 350 km above the equator? Hint: You don't need to know the mass of the space shuttle.
I need an explanation for how to start A and B because I dont understand how we have enough information with just the radius?
a.) To calculate the adjusted value of g at the equator due to centripetal acceleration, we need to consider that the centripetal acceleration is given by:
ac = (v^2) / r
Where v is the velocity and r is the radius of the Earth. At the equator, the velocity due to Earth's rotation is given by:
v = 2πr / T
Where T is the period of rotation, which is 24 hours.
Substituting for v in the centripetal acceleration formula:
ac = ((2πr / T)^2) / r
= (4π^2r) / T^2
Now, the adjusted value of g can be calculated by subtracting this centripetal acceleration from the standard gravitational acceleration, g:
g_adjusted = g - ac
Substituting the given values:
g_adjusted = 9.80 m/s^2 - ((4π^2 * 6380 km) / (24 hours)^2)
b.) The difference in g at the equator compared to the poles is indeed due to centripetal forces. At the poles, the centripetal acceleration is zero because there is no rotational motion. Therefore, the gravitational acceleration at the poles is equal to the standard value, g.
We use the standard value of 9.80 m/s^2 in calculations because it is an approximate average value for the acceleration due to gravity at sea level worldwide. Although g varies slightly depending on location and factors like altitude and local topography, the use of a standard value simplifies calculations and provides a close enough approximation for most applications.
c.) To find the value of g on board a space shuttle in a stable orbit at a height of 350 km above the equator, we can use the formula:
g_orbit = g_adjusted * (Re / (Re + h))^2
Where g_adjusted is the adjusted value of g at the equator, Re is the radius of the Earth, and h is the height above the equator.
Substituting the given values:
g_orbit = g_adjusted * (6380 km / (6380 km + 350 km))^2
Calculating this expression will give you the value of g on board the space shuttle in a stable orbit at a height of 350 km above the equator.
c)...in the orbit, folks are weightless.
a,b I will be happy to critique your thinking. YOu know velocity at the equator...r*2PI/24hrs