Question

The mass of a satellite is $M/81$ and radius is $R/4$ where $M$ and $R$ are the mass and radius of the planet. The distance between the surface of planet and its satellite will be at least greater then

• A. $1.25R$
• B. $12.5R$
• C. $10.5R$
• D. $5R$

Answer 1

The correct option is A $1.25R$

At $P$ the gravitational pull is same.

That is,

$\frac{GMm}{(R+x{)}^2}=\frac{GMm}{81(\frac{R}{4}+y{)}^2}$

$(R+x{)}^2=81(\frac{R}{4}+y{)}^2=\left[9\right(\frac{R}{4}+y){]}^2$

$R+x=9(\frac{R}{4}+y)$

Let $x+y=r$

$⟹R+(r-y)=9(\frac{R}{4}+y)$

$y=\frac{1}{10}[r-\frac{5}{4}R]$

Since $y>0$, $r>\frac{5}{4}R$

So the distance between the surface of planet and its satellite will be at least greater than $1.25R$