a mass of 5kg is first weighed on a balance at the top of a tower 20m high.the mass is then suspended from a fine wire 20m long and reweighted. Find the difference in weight.Assume that the radius of earth is 6400km mass of earth is 6*10^24 and G=6.67*10^(-11)
Answers
Answer:
m=5 kgm=5 kg is the mass that is initially hanigng
rE=6,400 km=6,400,000 mrE=6,400 km=6,400,000 m is the radius of the Earth
h=20 mh=20 m is the height of the tower
M=6 × 1024 kgM=6 × 1024 kg is the mass of the Earth
G=6.673 × 10−11 Nm2/kg2G=6.673 × 10−11 Nm2/kg2 is the universal gravitational constant
We can determine the weight of an object at any point by using Newton's law of gravitation:
W=GMmr2W=GMmr2
At the top of the tower, the distance between the mass and the center of the Earth is:
r1=6,400,000 m+20 m=6,400,020 mr1=6,400,000 m+20 m=6,400,020 m
When the mass is suspended from a 20 m20 m long wire, the mass can basically be treated as being on the ground. We can thus set r to just be the radius of the Earth:
r2=6,400,000 mr2=6,400,000 m
We can determine the difference in weight by subtracting the gravitational forces on the top and at the bottom of the tower:
ΔW=GMmr22−GMmr21ΔW=GMmr22−GMmr12
We substitute our values:
ΔW=(6.673 × 10−11 Nm2/kg2)(6 × 1024 kg)(5 kg)(6,400,000 m)2−(6.673 × 10−11 Nm2/kg2)(6 × 1024 kg)(5 kg)(6,400,020 m)2ΔW=(6.673 × 10−11 Nm2/kg2)(6 × 1024 kg)(5 kg)(6,400,000 m)2−(6.673 × 10−11 Nm2/kg2)(6 × 1024 kg)(5 kg)(6,400,020 m)2
We can simplify these terms: