Question

explain it plzzzz....​

Answer 1

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❏ Question:-

Refer to the questioner attachment,

❏ Solution:-

let,

the distance at apogee = $r_a$

the distance at perigee = $r_p$

and the mass of the body = m

Now we know that ,

$\longrightarrow \boxed{r_a=a(1-e)}$

$\longrightarrow \boxed{r_p=a(1+e)}$

where ,

e = eccentricity of the semi major axis .

a = length of the semi major axis.

Now , from the conservation theorem of Angular momentum .

$\longrightarrow \text{Angular momentum at apogee}= \text{Angular momentum at perigee}$

$\longrightarrow mV_ar_a=mV_pr_p$

$\longrightarrow V_ar_a=V_pr_p$

$\longrightarrow \dfrac{V_a}{V_p}=\dfrac{r_p}{r_a}$

$\longrightarrow \dfrac{V_a}{V_p}=\dfrac{\cancel{a}(1-e)}{\cancel{a}(1+e)}$

$\longrightarrow \boxed{\dfrac{V_a}{V_p}=\dfrac{(1-e)}{(1+e)}}$

Now,

$\longrightarrow \dfrac{K.E._p}{E.K._a}=\dfrac{\dfrac{1}{2}m{V^2}_p}{\dfrac{1}{2}m{V^2}_a}$

$\longrightarrow \dfrac{K.E._p}{E.K._a}=\dfrac{{V^2}_p}{{V^2}_a}$

$\longrightarrow \dfrac{K.E._p}{E.K._a}=\dfrac{(1+e)^2}{(1-e)^2}$

$\longrightarrow\large{\boxed{ \dfrac{K.E._p}{E.K._a}=\Big(\dfrac{1+e}{1-e}\Big)^2}}$

Refer to the attachment.,

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❏ Some Basic knowledge:-

➩ Apogee =

• Apogee is that point in a the elliptical orbit where the planet ( e.g, Earth) is at the farthest distance with respect to Sun .

➩ perigee =

• perigee is that point in a the elliptical orbit where the planet ( e.g, Earth) is at the nearest distance with respect to Sun .

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