For Kepler's laws of planetary motion, please see facts that matter.
The body will become weightless if the gravitational force mg on it is entirely used up in providing the centripetal acceleration for the rotation of the earth,
Then \(mg=\frac{m\theta^2}{R}=m\omega^2R\)
\(\omega^2=\frac{g}{R}=\frac{9.8}{6400\times10^3}\)
ω =1.237 x 10^-3 rad s^-1
If the earth rotates at this speed, the bodies on the equator will have no weight. At a latitude ф the apparent weight W_A is given by
\(W_A=mg(1-\frac{\omega^2R}{g}cos^2\phi)\)
Here, however, g = ω²R
Therefore,​​​​​​​ W_A= mg(1-cos² ф )
When  ф ​​​​​​​= 60⁰,cos ф ​​​​​​​= \(\frac{1}{2}\)
\(W_A=mg(1-\frac{1}{4})=\frac{3}{4}\times true\ weight\)
At poles, \(\phi=\frac{\pi}{2}\)
W_A mg = true weight
Thus, a body situated on the poles remains unaffected, whatever the speed of rotation of the earth.​​​​​​​

Acceleration due to gravity of the star \(g=\frac{GM}{R^2}\) .........(1)
Here M is the mass and R is the radius of the star.
The outward centrifugal force acting on a body of mass m at the equator of the star \(=\frac{mv^2}{R}=mR\omega^2\) .....(2)
From equation (i), the acceleration due to the gravity of the star
\(=\frac{6.67\times10^{-11}\times2.5\times2\times10^{30}}{(12\times10^3)^2}=2.316\times10^{12}m/s^2\)
\(\therefore\) Inward force due to gravity on a body of mass m
= m x 2.316 x 10¹² NFrom equation (ii), the outward centrifugal force = mRω²
\(=m\times(12\times10^3)\times(\frac{2\pi\times1.5}{-1})^2\)
= m x 1.06 x 10⁶ N
Since the inward force due to gravity on a body at the equator of the star is about 2.2 million times more than the outward centrifugal force, the body will remain stuck to the surface of the star.

Mass of Sun, M = 2 x 10³⁰ kg; Mass of Earth, m = 6 x 10²⁴ kg
Distance between Sun and Earth, r = 1.5 x 10¹¹ m

Let at the point P, the gravitational force on the rocket due to Earth 
= gravitational force on the rocket due to Sun
Let x = distance of the point P from the Earth
Then \(\frac{G_m}{x^2}=\frac{GM}{(r-x)^2}\)
\(\Rightarrow\frac{(r-x)^2}{x^2}=\frac{M}{m}=\frac{2\times10^{30}}{6\times10^{24}}=\frac{10^6}{3}\)
\(\frac{r-x}{x}=\frac{10^3}{\sqrt 3}\Rightarrow \frac{r}{x}=\frac{10^3}{\sqrt 3}+1\simeq\frac{10^3}{\sqrt 3}\)
\(x=\frac{\sqrt 3 r}{10^3}=\frac{1.732\times1.5\times10^{11}}{10^3}=2.6\times10^8m.\)

\(3.42\times 10^{ 5 }km\)

\(3.468\times 10^{ -10 }N\)

Initial velocity of the particle \(=15 \mathrm{~km} / \mathrm{s}\) Let its speed be V at intersteller space
\( \therefore\left(\frac{1}{2}\right) m\left[15 \times 10^3-v^2\right]=\int_R^{\infty} \frac{G M m}{x^2} d x \)
\( \Rightarrow\left(\frac{1}{2}\right) m\left[\left(15 \times 10^3\right)^2-v^2\right]=G M m\left[\frac{-1}{x}\right] \)
\( \Rightarrow\left(\frac{1}{2}\right) m\left[\left(225 \times 10^5\right)-v^2\right]=\frac{G M m}{R} \)
\( \Rightarrow 225 \times 10^5-v^2 \frac{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400 \times 10^3} \)
\( \Rightarrow v^2=225 \times 10^6-\frac{40.02}{32} \times 10^8 \)
\(=2.25 \times 10^8-1.2 \times 10^8 \)
\(=10^8(1.05) \)
\( \text { orv }=1.01 \times 10^4 \mathrm{~m} / \mathrm{s} \)
\( =10 \mathrm{~km} / \mathrm{s}\)

To solve this question, we use the Kepler’s third law. This states that the square of the period of the revolution of all the planets about the sun is directly proportional to the cube of the mean distance between the planets and the sun. This can be mathematically given as
T2 ∝ R3
⇒T2=kR3 where  T is the period of the revolution of a planet around the sun,  R is the mean radius of the planet to the sun.
Hence, we can write by comparison between two planets, that
\(\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}\)
We can use any planet with a known distance and period as the second planet. We choose earth.
\(\frac{T_N^2}{T_E^2}=\frac{R_N^3}{R_E^3}\) where the subscript N and E stands for Neptune and Earth respectively.
For earth, the period is 1 year. Hence, write that
\( \frac{165^2}{1^2}=\frac{R_N^3}{R_E^3} \)
\( \Rightarrow R_N^3=165^2 R_E^3\)
Hence, by finding the cube root of both sides, we have
\(R_N=\sqrt[3]{165^2} R_E\)
⇒RN=30RE
Radius of the earth is about 1.50×10¹¹m
Hence, RN=30(1.50×1011)=4.5×10¹²m

Acceleration due to gravity at depth d from the earth's surface is given by
\({ g }^{ ' }=g\left( 1-\frac { d }{ { R }_{ e } } \right) \)
Therefore, acceleration due to gravity decreases with increasing depth.

Gravitational attraction between two point objects are given as:
\(F = \dfrac{{G{M_1}{M_2}}}{{{r^2}}}\)…….(1)
Where,
F is the attractive force,
G is gravitational constant,
M₁ and M₂ are the masses of two objects,
r is the distance between the center of masses of the two objects.
The volume of a sphere of radius R is given by:
\(V = \dfrac{4}{3}\pi {R^3}\)……. (2)
Where,
V is the volume of the sphere,
R is the radius of the sphere.
Mass of an object with given density and volume:
\(M = \rho .V\)……. (3)
Where,
M is the mass of the object,
ρ is the density of the object,
Complete step by step solution:
Given:
The radius of the smaller sphere is R.
The radius of a larger sphere is 2R.
The density of both spheres is ρ.
The spheres are kept with their surface touching each other.
To find: Contact force between the spheres.
Step 1:
Use eq.(2) in eq.(3) to get the mass of the first sphere of R as:
\(M_1=\rho \times\left(\frac{4}{3} \pi R^3\right) \)
\(\therefore M_1=\frac{4}{3} \pi \rho R^3\)
Step 2:
Similarly, Use eq.(2) in eq.(3) to get the mass of the first sphere of $2 R$ as:
\(M_2=\rho \times\left(\frac{4}{3} \pi(2 R)^3\right) \)
\(\therefore M_2=\frac{32}{3} \pi \rho R^3\)
Step 3:
For a uniform sphere, its center of mass always stays at its center. As they are kept just in touch so the distance between their center is r=R+2R=3R. Now, substitute r and the values of  M₁ and M₂ obtained from eq.(4) and eq.(5) in eq.(1) to get the attractive force value as:
\(F=\frac{G \times\left(\frac{4}{3} \pi \rho R^3\right) \times\left(\frac{32}{3} \pi \rho R^3\right)}{(3 R)^2} \)
\( \therefore F=\frac{128}{81} G \pi^2 R^4 \rho^2 \)

The orbital velocity in a circular orbit close to Earth is: \(v=\sqrt{\mathrm{gR}} \text {. }\)
The velocity required to escape is \(v_e=\sqrt{2 g R}\)
\(v_e-v=(\sqrt{2}-1) \sqrt{g R}\)
Hence additional velocity required is: \(v_e-v=0.414 \times \sqrt{9.8 \times 6400 \times 10^3}=3278.71 \frac{\mathrm{m}}{\mathrm{s}}=3.278 \frac{\mathrm{km}}{\mathrm{s}}\)