NEET-XII-Physics
04: The Forces
- #7A satellite is projected vertically upwards from an earth station. At what height above the earth’s surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400 km.)Ans : Let h be the height, M be the Earth's mass, R be the Earth's radius and m be the satellite's mass
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Force on the satellite due to the earth when it is at the Earth's surface, `` {F}_{1}=\frac{\,\mathrm{\,GM \,}m}{{\,\mathrm{\,R \,}}^{2}}``
Force on the satellite due to the earth when it is at height h above the Earth's surface, `` {F}_{2}=\frac{\,\mathrm{\,GM \,}m}{{\left(\,\mathrm{\,R \,}\mathit{+}h\right)}^{2}}``
According to question, we have:
`` \frac{{F}_{1}}{{F}_{2}}=\frac{{\left(R+h\right)}^{2}}{{R}^{2}}``
`` \Rightarrow 2=\frac{{\left(R+h\right)}^{2}}{{R}^{2}}``
`` \,\mathrm{\,Taking \,}\,\mathrm{\,squareroot \,}\,\mathrm{\,on \,}\,\mathrm{\,both \,}\,\mathrm{\,sides \,},\,\mathrm{\,we \,}\,\mathrm{\,get \,}:``
`` \sqrt{2}=1+\frac{h}{R}``
`` \Rightarrow h=\left(\sqrt{2}-1\right)R``
`` =0.414\times 6400=2649.6\,\mathrm{\,km \,}\approx 2650\,\mathrm{\,km \,}``