Physics Gravitation

Yes. Escape speed from a planet or celestial body is the same for all objects. It is irrespective of their mass. This is because the gravitational force and potential energy depend on the mass of the celestial body and distance. The object's own mass cancels out when deriving escape speed. 

The escape speed from the Earth's surface is approximately 11.2 km/s. This means that the object must travel at least at this speed in the upwards direction to entirely escape Earth's gravitational pull without falling back.  

Escape velocity or escape speed is the minimum velocity required for an object to move away from the gravitational pull of a celestial body or planet. This velocity for escape does not require any additional propulsion. 

$g_{\left(ath\right)}=g_{\left(atdepth\alpha h\right)}$ h << R

$\Rightarrow g\left(1-\frac{2h}{R}\right)=g\left(1-\frac{\alpha h}{R}\right)$

$\begin{array}{l}1-\frac{2h}{R}=1-\frac{\alpha h}{R}\Rightarrow \frac{2h}{R}=\alpha \frac{h}{R}\\ \alpha =2\end{array}$

 $T^2\alpha R^3$

$\begin{array}{l}\Rightarrow \frac{{T_1}^2}{{T_2}^2}=\frac{{R_1}^3}{{R_2}^3}\\ \Rightarrow \frac{{T_1}^2}{{T_2}^2}=\frac{R^3}{{\left(3R\right)}^3}\\ T_2=3\sqrt[]{3}years\end{array}$

According to question, we can write

$d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2*125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

According to question, we can write

$d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2*125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

Yes, a body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, $f_g$ = $\frac{GMm}{R^2}$ , where M = mass of the star = 2.5 $*2*{10}^{30}$ = 5 $*{10}^{30}$ kg

M = mass of the body, R = radius of the star = 12km = 1.2 $*{10}^{4}m$

$f_g$ = $\frac{6.67*{10}^{-11}*5*{10}^{30}*m}{{{(1.2*{10}^4)}^2}^{}}=$ 2.31 $*{10}^{12}mN$

Centrifugal force $f_c$ = mr ${\omega }^2$ where $\omega =$ angular speed = 2 $\pi \gamma$ and angular frequency $\gamma$ = 1.2 rev/s

$f_c=$ mR( ${2\pi \gamma )}^2$ = m $*$ (1.2 $*{10}^4)*2*3.14*1.2*(2*3.14*1.2)$ = 6.81 $*{10}^{5}mN$

Since $f_g>f_c$ , the body will remain stuck to the surface of the star.