Physics help(gravitation)
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1) Certain neutron stars(extremely dense stars) are believed to be rotating at about one revolution per second. If such a star has a radius of 20 km, what must its minimum mass so that objects on its surface will still be attracted to the star and not be "thrown off" by the rapid rotation? Ans: 4.7 X 10^24kg
2) The asteroid Eros, one of the many minor planets that orbit the sun in the region between Mars and Jupiter, has a radius of 7.0km and a mass of 5.0 X 10^15 kg.
a) if you were standing on Eros, could you (giving reasons, ignoring rotation of asteroid),
i) lift a 2000 kg pickup truck?
ii) Push it?
Thanks for those who can help
very blur in physics 
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1) For object on surface not to be thrown off, gravitational attraction must provide the necessary centripetal force. Hence,
mrw^2= GMm/r^2
where m = arbitrary mass of an object on surface, M is mass of planet, r = radius of planet = distance between center of planet to object on surface, G = gravitional constant
mrw^2 = GMm/r^2
r^3 x w^2/G = M
M = 4.7 X 10^24kg
2i) W = mg where W = weight of truck = force needed to lift truck where g = gravitational acceleration of truck on surface, peculiar to the mass of the planet
Since gravitational force = GMm/r^2 = mg, where r = radius of planet = distance between center of mass of planet and object on surface of planet
g = GM/r^2 = ( 6.67 x 10^-11 x 5 x 10^15 ) / (7 x 10^3)^2 = 6.8 x 10^-3
Hence, force needed to lift truck = mg = 2000 x 6.8 x 10^-3 = 13. 6 N.
For most adults, 13.6 N is not a problem, so ans is can lift the truck.
ii) ? -
I also don't fully understand this questionOriginally posted by le cruz:ii) ? -
haha for ii) i meant pushing the same 2000kg truck .. forgot to include in..ps
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as for the second part, pushing it will depend alot on the friction wor....Originally posted by le cruz:1) For object on surface not to be thrown off, gravitational attraction must provide the necessary centripetal force. Hence,
mrw^2= GMm/r^2
where m = arbitrary mass of an object on surface, M is mass of planet, r = radius of planet = distance between center of planet to object on surface, G = gravitional constant
mrw^2 = GMm/r^2
r^3 x w^2/G = m
m = 4.7 X 10^24kg
2i) W = mg where W = weight of truck = force needed to lift truck where g = gravitational acceleration of truck on surface, peculiar to the mass of the planet
Since gravitational force = GMm/r^2 = mg, where r = radius of planet = distance between center of mass of planet and object on surface of planet
g = GM/r^2 = ( 6.67 x 10^-11 x 5 x 10^15 ) / (7 x 10^3)^2 = 6.8 x 10^-3
Hence, force needed to lift truck = mg = 2000 x 6.8 x 10^-3 = 13. 6 N.
For most adults, 13.6 N is not a problem, so ans is can lift the truck.
13.6 N is like less than 2 kg on earth
not only most adults....a kid can lift it up=p
ii) ?
coz (F-f)=dm/dt * dv/dt (thus the formula F=ma)
anyways to move it u will need a resultant force. if ur sayin pushing it horizontally on the surface of the planet, then u will need juz F>f where F is the force applied n f is friction. -
Any more details left out?Originally posted by lavastar:haha for ii) i meant pushing the same 2000kg truck .. forgot to include in..ps -
apparently no according to the question i was given.Originally posted by eagle:Any more details left out?