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.
The value of g would increase if the compound pendulum is taken nearer to the center of the Earth. This is because gravity is stronger closer to the Earth's surface. Conversely, if the compound pendulum is moved further away from the center of the Earth, the value of g would decrease.
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 decreases when you go deeper within the Earth. That is because part of the Earth will be above you, and therefore pull you up. your wrong my friend: in Thermodynamics by Cenjel: It is interesting to note that at locations below sea level, the value of g increases with distance from the sea level, reaches a maximum at about 4500 m, and then starts decreasing. (What do you think the value of g is at the center of the earth?)
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.
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.
The value of g would increase if the compound pendulum is taken nearer to the center of the Earth. This is because gravity is stronger closer to the Earth's surface. Conversely, if the compound pendulum is moved further away from the center of the Earth, the value of g would decrease.
Yes, the maximum value of acceleration due to gravity (g) is at the equator because the centrifugal force due to the Earth's rotation reduces the effective gravitational force. This results in a maximum g value at the equator compared to other latitudes.
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
at the poles
The weight is due to the force exerted on the body by the gravitational pull by the earth. If no such gravitational pull on the body, then the body becomes weightless. In free space, very far away from the earth the astronauts experience weightlessness. So if M is the mass of the body, then Mg is the weight of the body. Hence weight depends on the value of 'g'. The value of g is the least at the equatorial region. It will be maximum at the poles. As we go at higher altitudes g value decreases. Same way as we go towards the centre of the earth ie as depth increases, g value decreases.
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.
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.
g
The gravitational force decreases when you go deeper within the Earth. That is because part of the Earth will be above you, and therefore pull you up. your wrong my friend: in Thermodynamics by Cenjel: It is interesting to note that at locations below sea level, the value of g increases with distance from the sea level, reaches a maximum at about 4500 m, and then starts decreasing. (What do you think the value of g is at the center of the earth?)
As you move deeper into the Earth, the value of acceleration due to gravity (g) decreases slightly. This is because the mass directly below you is pulling you down, while the mass above you is also pulling you up. The net effect of these opposing forces is a slight decrease in the value of g as you move deeper into the Earth.
Zero.
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.