Question

The change in value of acceleration of earth towards sun, when the moon comes from the position of solar eclipse to the position on the other side of earth in line with sun is . Take mass of moon equal to
$7.36 \times 10^{22} kg$
and the orbital radius of moon is
$3.8 \times {10}^{8} m$
It is so urgent please answer it

Answer 1

Ans. is 6.8 x 10^(-5) m/s^2
solution is in the pic.... :)

Answer 2

Change in acceleration of earth towards sun from solar eclipse to lunar eclipse period is $6.799 \times 10^{-5} \ \mathrm{m} / \mathrm{s}^{2}$

Explanation:

During solar eclipse, the sun and moon will be on the same side of the earth and during lunar eclipse, the moon and the sun will be exactly on the earth’s opposite sides.

So the gravitational force acting on earth due to sun will add up with the gravitational force acting on earth due to moon during solar eclipse and the gravitational forces will subtract with each other during lunar eclipse.

Let $F_s$ be the gravitational force between earth and sun and $F_L$ be the gravitational force between earth and moon.

Also, the acceleration of earth during solar and lunar eclipse can be derived from Newton’s second law, i.e.,

$F=m \cdot a \rightarrow(1)$

During solar eclipse, eqn (1) will be  

$m \cdot a_{s}=F_{s}+F_{L} \rightarrow(2)$

During lunar eclipse, eqn (1) will be  

$m \cdot a_{L}=F_{s}-F_{L} \rightarrow(3)$

Here, a_s and a_L are the acceleration of earth towards sun during solar and lunar eclipse respectively and m is the mass of earth.

So the change in acceleration of earth toward sun from solar to lunar eclipse period, can be found by subtracting eqn (2) from eqn (3)  

$m\left(a_{s}-a_{L}\right)=F_{s}+F_{L}-F_{s}+F_{L}$

$\therefore m\left(a_{s}-a_{L}\right)=2 F_{L} \rightarrow(4)$

Here,

$F_{L}=G \frac{m \cdot M_{L}}{D^{2}}$

Where G is the gravitational constant, m is the mass of earth, ML is the mass of moon, and D is the distance between moon and earth.

D is equal to the orbital radius of moon.

So the eqn (4) becomes,

$\left(a_{s}-a_{L}\right)=2 G \frac{M_{L}}{D^{2}} \rightarrow(5)$

Now, substitute the given values in eqn (5),

$M_{L}=7.36 \times 10^{22} \ \mathrm{kg}$

$D=3.8 \times 10^{8} \ m$

Gravitational constant value, $G=6.67 \times 10^{-11} \ N . m^{2} . K g^{-2}$

$\left(a_{s}-a_{L}\right)=2 \times 6.67 \times 10^{-11} \times \frac{7.36 \times 10^{22}}{\left(3.8 \times 10^{8}\right)^{2}}$

$\left(a_{s}-a_{L}\right)=6.799 \times 10^{-11+22-16}$

$\left(a_{s}-a_{L}\right)=6.799 \times 10^{-5} \ \mathrm{m} / \mathrm{s}^{2}$