Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The figure shows the elliptical orbit of a planet $m$ about the sun $S$ . The shaded area $SCD$ is twice the shaded area $SAB$ . If $t_1$ is the time for the planet to move from $C$ to $D$ and $t_2$ is the time to move from $A$ to $B$ , then:

$t_1=3t_2$

$t_1=4t_2$

$t_1=2t_2$

$t_1=t_2$

Step-by-Step Solution

Kepler's Second Law: The Law of Areas states that the line joining the sun to the planet sweeps out equal areas in equal intervals of time . This implies that the areal velocity ( $\frac{\Delta A}{\Delta t}$ ) is constant. Consequently, the area swept ( $A$ ) is directly proportional to the time taken ( $t$ ). $A \propto t \quad \text{or} \quad \frac{A_1}{t_1} = \frac{A_2}{t_2}$
Given Data:

Area corresponding to time $t_1$ (path C to D) is $A_{SCD}$ .
Area corresponding to time $t_2$ (path A to B) is $A_{SAB}$ .
Condition: $A_{SCD} = 2 \times A_{SAB}$ .

Calculation: Substitute the condition into the proportionality equation: $\frac{A_{SCD}}{t_1} = \frac{A_{SAB}}{t_2}$ $\frac{2 A_{SAB}}{t_1} = \frac{A_{SAB}}{t_2}$ $\frac{2}{t_1} = \frac{1}{t_2} \implies t_1 = 2t_2$

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