Two satellites of masses m and 1.5 m are revolving around the earth with different speeds in two circular orbits of heights R_E and 2 R_E respectively, where R_E is the radius of the earth. The ratio of the minimum and maximum gravitational forces on the earth due to the two satellites is

Correct answer: (D) 1: 5. Maximum force, F_max =F_1+F_2 =fracCM_m (2 R_E )^2+frac1.5 GM_m (3 R_E )^2=5/12 fracGM_mR_E^2 Minimum force, F_min =F_1-F_2 =fracGM_m (2 R_E )^2-frac1.5 CM_m (3 R_E )^2=1/12 · fracGM_mR_E^2 ∴ fracF_min F_max =fracGM_m12 R_E^2 × frac12 R_E^25 GM_m=1/5=1: 5

Q133 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics

Two satellites of masses m and 1.5 m are revolving around the earth with different speeds in two circular orbits of heights $\mathrm{R}_E$ and $2 \mathrm{R}_E$ respectively, where $\mathrm{R}_E$ is the radius of the earth. The ratio of the minimum and maximum gravitational forces on the earth due to the two satellites is

A
$2: 5$
B
$2: 3$
C
$1: 2$
Correct answer
D
$1: 5$

Solution

Maximum force, $\mathrm{F}_{\max }=\mathrm{F}_1+\mathrm{F}_2$ $=\frac{\mathrm{CM}_{\mathrm{m}}}{\left(2 \mathrm{R}_{\mathrm{E}}\right)^2}+\frac{1.5 \mathrm{GM}_{\mathrm{m}}}{\left(3 \mathrm{R}_{\mathrm{E}}\right)^2}=\frac{5}{12} \frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{E}}^2}$ $\begin{aligned} & \text { Minimum force, } \mathrm{F}_{\min }=\mathrm{F}_1-\mathrm{F}_2 \\ & =\frac{\mathrm{GM}_{\mathrm{m}}}{\left(2 \mathrm{R}_{\mathrm{E}}\right)^2}-\frac{1.5 \mathrm{CM}_{\mathrm{m}}}{\left(3 \mathrm{R}_{\mathrm{E}}\right)^2}=\frac{1}{12} \cdot \frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{E}}^2}\end{aligned}$ $\therefore \frac{\mathrm{F}_{\min }}{\mathrm{F}_{\max }}=\frac{\mathrm{GM}_{\mathrm{m}}}{12 \mathrm{R}_{\mathrm{E}}^2} \times \frac{12 \mathrm{R}_{\mathrm{E}}^2}{5 \mathrm{GM}_{\mathrm{m}}}=\frac{1}{5}=1: 5$

Hence, the correct answer is option: 'D'.

Solution

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