Tltr: The quesion d makes no sence to me and I dont know what my teacher mean. 1. The problem statement, all variables and given/known data The Earth turns once around its axis in 24.0 hours. We will assume that it is perfectly spherical, with radius 6400 km. The mass of the Earth is taken to be 6.00 × 1024 kg. The gravitational acceleration on the surface is taken to be g = 9.80 m/s2 . a) What is the centripetal acceleration of a person at the equator? How big a fraction of the gravitational acceleration does this correspond to? Remember to draw a sketch of the situation. b) How fast would the Earth have to turn for the centripetal acceleration to be exactly equal to gravity? Give the answer in revolutions per day. Remember to draw a sketch of the situation. c) Define θ to be the latitude, so that θ = 0 corresponds to the equator, and θ = 90◦ is the North Pole. What is the centripetal acceleration of a person standing on the surface at a given value of θ? Remember to draw a sketch of the situation. d) Taking into account that the centripetal acceleration is towards the rotation axis, and gravity is towards the centre of the Earth, how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)? Remember to draw a sketch of the situation. e) Using the force of gravity |Fg| = G (Mm)/r^2 , where M is the mass of the Earth, m the mass of the orbiting object. Find the required speed vorbit, as a function of r, so that the object performs uniform circular motion around the Earth, under the influence of gravity. How long does it take to go round once in an orbit 900 km above ground? Remember to draw a sketch of the situation. 2. Relevant equations a=v^2/r v = (2*pi*r)/T , where T is the time it takes for the earth to pass one round around it's own axis (24*60*60)s 3. The attempt at a solution A) I used a=v^2/r and found it to be 0.0338 m/s^2 9.8/0.0338 = 290 thus the acceleration is 1/290 of g B) By setting 9.8 = (V^2)/r and solving it for v i got 7920 m/s Bearing in mind that T=(2*pi*r)/v I got 5078 sec. or 84.6 min. (24*60)/84.6=17 rounds per 24 hours C) This one is difficult to explain here but I'll try my best. Thinking that the radius for the person gets smaller as one gets further north (because the earth is smaller there) I found that the "new" radius is r*cos(tetta) where r is the original radius Putting this into the a=(v^2)/r gives this witch I hope is the correct one : (4*pi^2*r*cos(tetta))/T^2 I've allso solved e but the problem is for me d "how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)?" I thought that i could decompose g in x- and y-direction because as the task sais, a is in horizontal direction and g is in both x and y since it towards the center of earth. I tried to find then when the x-component of g (gcos(tetta)) is equal to my answer from task c. This resulted in the angle dissaparing completly from my equation and so it must be wrong. Quite francly I don't understand the question. How can g produce the necessary centripetal acceleration when that should come from the movent of the earth around its own axis. Those two accelerations has nothing to do with eachother as far as I can understand. Also what is the necessary centripital acceleration?