Physics Gravitation

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2*3.14}{\sqrt[]{6.67*10^{-11}*6*10^{24}}}\left[{\left(8*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$F_{1}cos45°+F_{2}cos45°+F_3=\frac{m{v}^2}{r}$

$\Rightarrow \frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(2r\right)}^2}=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt[]{\frac{Gm}{4r}\left(2\sqrt[]{2}+1\right)}$

Putting m = 1 kg and r = 1 m,

$v=\frac{1}{2}\sqrt[]{G\left(1+2\sqrt[]{2}\right)}$

At each point in the orbit, there is a variation in the kinetic and potential energy of the satellite. If one increases, the other one decreases. Similarly, if the other one increases, the first one decreases. But their amount of increment or decrement is such that the total sum i.e. mechanical energy of the satellite remains the same. This in turn allows free movement.

This is because the satellite doesn't have any other external third force acting upon it. Its original total energy comprises kinetic and potential energy, which remains constant since the gravity has zero influence over the satellite.

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

$g\propto r0\le r\le R$

$g\propto \frac{1}{r^2}r>R$

The minus sign in the formula indicates that:

• A gravitational force exists.
• The satellite cannot leave the orbit by itself.
• A minimum threshold of energy will be required to move the satellite from its existing orbit.

According to kepler's third law of time period –

  ${\left(\frac{T_A}{T_B}\right)}^2={\left(\frac{r_A}{r_B}\right)}^3\Rightarrow \frac{r_A}{r_B}=4^{1/3}$  

$\Rightarrow r_4^3=4r_B^3$  

Kindly go through the solution 

$b=a\sqrt[]{1-e^2}$

$\frac{dA}{dt}=\frac{L}{2m}$

$\frac{A}{T}=\frac{L}{2m}$

$T=\frac{2mA}{L}=\frac{2m\pi ab}{L}=\frac{2m\pi a^2\sqrt[]{1-e^2}}{L}$