Why do rocket choose the equator as a starting point to blast off?
Some books say the equator is the furthest from the centre of Earth
so it has the greatest rotational velocity.
But according to the equation, F= GMm / R^2
where r=the distance between the centre of earth and the rocket
As the distance between the centre of earth and the rocket increases,
the gravitational force increases
which reduces the kinetic energy
( w= fs -----> f= w/s ----->f= 0.5mv^2 /s )
and the velocity decreases
Could someone tell me if there is something wrong?
Well think about it - we assume the Earth is a sphere, which, by definition, has all points on the surface an equal distance from the centre so NO - The equator is NOT the furthest from the centre of the Earth.
But yes, the equaor does have the greatest rotational velocity - do you have a model globe at home that spins? if you do it is easy to see why this is - the Earth rotates about an axis - this axis is a line through the North and South Poles, so the motion is circular (very easy to see witht he globe) - we know it takes (approximately) 24 hours to rotate once completely, however what is 'one rotation'? at the equator, one rotation is a circular path, with circumference of this path (ie distance travlled) equal to c=2(pi)r - this is the case with ALL points on the Earth, however the 'r' here is NOT the radius of the Earth in all cases - it is the distance from the 'axis of rotation ' (the line joining the North and South poles) to that point - this, at the North and South poles is 0 [which is why the sun never rises or sets in Antarctica for several months of the year] - and it continues to increase until the equator, at which poinr 'r' is equal to the radius of the Earth
It is clear, since speed = distance / time, the time to complete 'one rotation' is constant no matter where you are on Earth - 24 hours, so speed = distance/(24*60*60 [secs]) - so rotational veloicty is a function of 'distance' - which changes form point to point. Remember, we said distance traveled was 2(pi)r, where 'r' is the [perpendicular] distance between the axis of rotation [line joining N and S poles] and that point, and so
rotational velocity = distance/(24*60*60 [secs]) = (2(pi)r)/(24*60*60 [secs])
and so we can see, rotational velocity is DIRECTLY PROPORTIONAL to 'r' [which means the greater 'r' is, the greater rotational velocity is], and we know 'r' is greatest at the equator, so the equator has the greatest rotational velocity.
For a less mathematical approach as to why the rotational velocity is greatest at the equator, think about it - at the equator, a greater distance is covered by the rotation of the Earth in the same amount of time [as everywhere else on the Earth], which can only be done if the seped of the Earth's rotation is faster at that point
Hope that helps
Also (1), when speaking about 'rotational' and 'orbital' velocities of the Earth and why they help 'boost' a rocket - always say 'relative to the Sun' - remember, the rotational motion of the Earth adds 1700 km/h to the speed of a rocket launched towards teh East at the equator RELATIVE TO THE SUN - this is important, as relative to the Earth [which is travelling in rotation at 1700 km/h towards the East] the rocket is not going any faster than we perceieve it to be.
Also (2), 'escape velocity' is constant everywhere on the Earth, if we assume the Earth is a perfect spehere (which is reasonable) so that is NOT the reason we launch rockets to the East from the Equator.
Also (3), 'escape velocity' is really irrelevant when launching rockets - remember escape velocity deals with the minimum speed which is needed to launch an UNPROPELLED object into space, and is approx 11.2 km/s - never in reality do we send rockets unpropelled - we can 'escape' the gravity of the Earth (or rather the 'effectivenees of the Earth-object mutual gravitation', as we can never escape gravity) at any speed we wish - 1 m/s if we want - as long as we are being propelled.
