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a) Is the acceleration due to the gravity of earth ‘g’ a constant? Discuss.
b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4\ \times10^{22\ } kg and a radius of 1.74\times10^6 m. Which satellite do you think it could be?
Solution
Transcript
a) No, the value of acceleration due to gravity is no constant at all the places on the surface of the earth. This is because the radius of the earth is minimum at the poles while the maximum at the equator. The value of g is maximum at the poles and minimum at the equator. The value of g increases as we go from the centre of the earth towards the poles and the value of g decreases as we go inside the earth. b) Acceleration due to gravity,
g = G M/R2
Mass, M = 7.4\ \times10^{22\ } kg
Radius, R = 1.74\times10^6 m
Gravitational constant, G = 6.7\times10^{-11}\ Nm^2/kg^2
Therefore, substituting the values we get g = 1.637 m/s^2
"hey kids welcome to lido q a video i am vineet your lido tutor bringing you this question on your screen now question is is the acceleration due to gravity of earth g a constant discuss so let us answer the first question discuss the first question and then answer the second question so the clear answer for this is no the acceleration due to gravity of earth earth is not is not constant constant at all the places on the surface of the earth all the places on the surface of the earth why is this this is because the radius of the earth is minimum this is because the radius of the earth is minimum at the poles and poles so i'll have to continue this here and maximum near the equator right now how does that impact the uh acceleration due to gravity this impacts the acceleration due to gravity because this actually impacts the acceleration due to gravity because the value of g if is inversely proportional to the radius or the distance right so that that's how so the value of g is maximum at the poles and approaches zero as we move to the center or the core of the earth right now this is party let us look let us all part b now now in part b it is given to us that mass of earth is mass of the satellite is 7.4 into 10 to the power 22 kg and radius of the satellite is 1.74 into 10 to the power 6 meter right and g we know is equal to 6.6 7 into 10 to the power minus 11 newton meter square per second square now g is given by we know that g is equal to gravitational constant into mass by radius square so substituting the values and solving we get g is equal to 1.637 meter per second square isn't that easy guys if you still have a doubt though please leave a comment below do like the video and subscribe to our channel and see you in our next video until then bye guys keep learning"
a) Is the acceleration due to the gravity of earth ‘g’ a constant? Discuss.
b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4\ \times10^{22\ } kg and a radius of 1.74\times10^6 m. Which satellite do you think it could be?
Solution
Transcript
a) No, the value of acceleration due to gravity is no constant at all the places on the surface of the earth. This is because the radius of the earth is minimum at the poles while the maximum at the equator. The value of g is maximum at the poles and minimum at the equator. The value of g increases as we go from the centre of the earth towards the poles and the value of g decreases as we go inside the earth. b) Acceleration due to gravity,
g = G M/R2
Mass, M = 7.4\ \times10^{22\ } kg
Radius, R = 1.74\times10^6 m
Gravitational constant, G = 6.7\times10^{-11}\ Nm^2/kg^2
Therefore, substituting the values we get g = 1.637 m/s^2
"hey kids welcome to lido q a video i am vineet your lido tutor bringing you this question on your screen now question is is the acceleration due to gravity of earth g a constant discuss so let us answer the first question discuss the first question and then answer the second question so the clear answer for this is no the acceleration due to gravity of earth earth is not is not constant constant at all the places on the surface of the earth all the places on the surface of the earth why is this this is because the radius of the earth is minimum this is because the radius of the earth is minimum at the poles and poles so i'll have to continue this here and maximum near the equator right now how does that impact the uh acceleration due to gravity this impacts the acceleration due to gravity because this actually impacts the acceleration due to gravity because the value of g if is inversely proportional to the radius or the distance right so that that's how so the value of g is maximum at the poles and approaches zero as we move to the center or the core of the earth right now this is party let us look let us all part b now now in part b it is given to us that mass of earth is mass of the satellite is 7.4 into 10 to the power 22 kg and radius of the satellite is 1.74 into 10 to the power 6 meter right and g we know is equal to 6.6 7 into 10 to the power minus 11 newton meter square per second square now g is given by we know that g is equal to gravitational constant into mass by radius square so substituting the values and solving we get g is equal to 1.637 meter per second square isn't that easy guys if you still have a doubt though please leave a comment below do like the video and subscribe to our channel and see you in our next video until then bye guys keep learning"
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