A) 23 N
B) 25 N
C) 21 N
D) 27 N
Correct Answer: D
Solution :
As we know that \[g=\frac{Gm}{{{(R+h)}^{2}}}\] At, \[h=36000\,km\] \[{{g}_{1}}=\frac{GM}{{{(36000+6400)}^{2}}}=\frac{GM}{{{(42400)}^{2}}}\] Also, at the surface h = 0, then \[{{g}_{2}}=\frac{GM}{{{(6400)}^{2}}}\] Now, \[{{g}_{1}}/{{g}_{2}}=\frac{6400\times 6400}{42400\times 42400}=\frac{256}{106\times 106}\] \[=0.0227\] \[(\because \,{{g}_{2}}=9.8)\] \[{{g}_{1}}=0.0227\times 9.8=0.223\] Thus, weight of 120 kg equipment placed in satellite \[=m{{g}_{1}}\] \[=120\times 0.223=26.79N\]
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