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What will happen to the value of g if the compound pendulum is taken nearer to the centre of earth?
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.
What is the value of g at the centre of earth in detail?
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.
What is the value of g at depth of earth?
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?)
What will be the time period of a simple pendulum at the centre of 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.
What is the value of 'g' if the earth stops rotating?
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.
What will happen to the value of g if the compound pendulum is taken nearer to the centre of earth?
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.
Is the maximum value g at equator?
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.
Where the gravitational field strength is maximum?
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
The weight of an object due to?
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.
Where is maximum value of g?
at the poles
What is the earth's gravity in terms of acceleration?
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.
What is the value of g at the centre of earth in detail?
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.
Who believed in 340BC that the earth was the centre of your solar system?
g
What is the value of g at depth of earth?
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?)
What is the effect of depth on the earth on the value of g?
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.
What will be the time period of a simple pendulum at the centre of 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.
What height from the centre of earth the value of g is half of its value on the surface of earth?
gravity follows an inverse squared law. at twice the distance it would be a quarter of G. 1/g2 = 1/2. so the distance would be the square root of 2 (or 1.4). 1.4 x earths radius of 6366 km would be 9003 km
