At the center of a sphere, for every dot of mass attracting you toward it, there's
another dot of mass at exactly the same distance from you in exactly the opposite
direction, attracting you toward it and canceling out the first one.
At the center, for every particle of mass in the earth to attract you gravitationally, there's an equal particle with the same mass, located at the same distance from you in exactly the opposite direction, balancing out the force toward the first one. All of this is a theoretically ideal case. It assumes that the earth is a perfect sphere, with the same distribution of mass in every direction from the center. The real earth is lumpy, oddly shaped, and bumpy, so the acceleration of gravity wouldn't be exactly zero at the center.
In a zero-gravity environment, you would not experience the normal effects of G-force caused by ambient gravity. In a zero-g environment, you are essentially in free-fall, so you are weightless and there is no force acting on you to create a sensation of gravity.
a gravitational force is a force of attraction , which attracts the all organisms and things , it also attracts the things in space [universe for where the gravitational field is there] for suppose our earth is a planet which have gravity , a sun[star] have more gravity in solar system. Improved answer: Gravitational force on a body brings the weight of the body. This is given by the expression W = mg. m-the mass of the body. g-the acceleration due to gravity. When this g gets affected then gravitational force varies. g is affected by so many factors. i) the altitu. de as we go away from the centre of the earth, g decreases ii) The depth. As we go towards the centre of the earth once again g decreases iii) Due to rotation of the earth the centrifugal acceleration would decrease the value of g, the most along the equator. iv) g could be proved to be GM/R2. G the universal gravitational constant M-mass of the earth R- radius of the earth. Hence due to non even spherical shape of the earth g value varies v) if a body gets immersed in a liquid then the weight is affected by the buoyant force. So the weight is reduced.
value of acceleration due to gravity is maximum at the surface of earth. So the gravitational field strength. as g'=g(1-d/R) at surface d=R so d=R so g'=g at earth's centre g=0. Its value decrease with decrease or increase in height. as: g'=g(1-2h/R) ......for height h and g'=g(1-d/R) .....for depth d
Any time a person falls, they experience reduced gravity for a short time. If they fall a greater distance, the more likely they will experience something close to zero G. Sky diving, riding a roller coaster/ amusement park ride with a free fall component can get close to zero g. If you've been in a plane that hit turbulence and dropped quite a ways, you may have experienced zero gravity. Point is, zero is experienced during free fall, which is exactly what the astronauts are doing when they are in orbit.
9.8 m/s2 ---------------------- Yes this is the average value of acceleration due to gravity near by the surface of the earth. As we go higher and higher level this g value decreases and becomes almost negligible. Same way as we go deeper and deeper the g value decreases and at the centre of the earth its value becomes zero.
No, at the center of the Earth, it would be zero. That's because the gravitation of different parts of the Earth, in different directions, would cancel.
g = 9.81 m/s2 = 32.2 ft/s2the above is a wrong answer. that's g on the surface of the earth.g varies and actually decreases as we move up or down the surface. g at the centre of any spherical body due to it's own mass is 0. So g at the centre of earth is zero.
If the Earth were to stop rotating, the value of 'g' (acceleration due to gravity) would remain approximately the same at the Earth's surface. The rotation of the Earth does not significantly affect the gravitational pull experienced on the surface.
Zero.
g
The value of acceleration due to gravity (g) at the center of Earth is theoretically zero. This is because the mass surrounding the center exerts equal gravitational force in all directions, effectively canceling each other out at the center point.
The gravitational force between the earth and a body at the center of the earth would be 0 Newtons or 0 lbf. F=G (m1*m2)/r^2 r = zero if the center of the body is at the center of the earth
The value of acceleration due to gravity 'g' at the center of the Earth is theoretically zero because the mass of the Earth surrounds an object equally in all directions, resulting in a net gravitational force of zero at the center.
Gravitational acceleration increases ,as distance from the centre decreases g=GM/s2 where, G=gravitational constant(6.67by 10^-11) M=mass of any pulling body(earth) s=distance from centre Though not infinite for the earth(is infinite for black hole) but the value of 'g' and hence, gravity is very high at the centre of the earth(6000 km below us), so a pendulum will swing very very fast at centre and its time period will be nearly zero T2=4(3.142)2l/g ================================= The centre of the Earth is occupied by solid material. A pendulum could not swing at that location. Also at the very centre of the earth all the mass is uniformly outside your location - there is no effective down position, so if you created a void to swing the pendulum it would not swing.
There will be no gravity. Objects will start floating.
At the center, for every particle of mass in the earth to attract you gravitationally, there's an equal particle with the same mass, located at the same distance from you in exactly the opposite direction, balancing out the force toward the first one. All of this is a theoretically ideal case. It assumes that the earth is a perfect sphere, with the same distribution of mass in every direction from the center. The real earth is lumpy, oddly shaped, and bumpy, so the acceleration of gravity wouldn't be exactly zero at the center.