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Headache is due to mental strain.It will persist whether a person in an astronaut in space or he is on earth.It means headache will have the same effect on the astronaut in space as on a person on earth.

The weight of an object below surface of the earth is zero.

Parking orbit that orbit in which the period of revolution of a satellite is equal to the period of rotation of the earth about its axis.

\(F=G\frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } }\)
\( \\ =6.67\times 10^{ -8 })\left( \frac { 1\times 1 }{ 1^{ 2 } } \right) dyne\)
\(\\ =6.67\times 10^{ -8 }\quad dyne=\frac { 6.67\times 10^{ -8 } }{ 980 } \)
\(=7\times 10^{ -11 }g-wt\)

No,we cannot determined the mass of a satellite by measuring its time period

Step 1: Given
1. The equal mass of the particles is m
2. The radius of the circle is R.
Step 2: Formula used
1. Gravitational force \(=\frac{G M m}{R^2}\)
2. Centrifugal force \(=m \omega^2 R\)
Step 3: Solution
Let the speed of rotation of each particle be v.
By equating gravitational force with centrifugal force
\(\frac{G m^2}{(2 R)^2}=m \omega^2 R \)
\(G=\text { Gravitational constant }\)
2R =  The distance between the center of the two particles.
\(\omega=\) Angular velocity of the particles
Now,
\(\Rightarrow  \frac{G m^2}{4 R^2}=m \omega^2 R \)
\(\Rightarrow  \frac{G m^2}{4 R^2 \cdot m R}=\omega^2 \)
\(\Rightarrow \frac{G m}{4 R^3}=\omega^2 \)
\(\Rightarrow  \omega =\sqrt{\frac{G m}{4 R^3}}\)
From the relation between angular velocity and linear velocity
\( v=\omega R \)
\( \Rightarrow v=\sqrt{\frac{G m}{4 R^3} \times R} \)
\( \Rightarrow v=\sqrt{\frac{G m R^2}{4 R^3}} \)
\( \Rightarrow v=\sqrt{\frac{G m}{4 R}}\)
[Putting value of \(\omega\) from (1)]
Hence the required answer is \(v=\sqrt{\frac{G m}{4 R}}\).

No, they are applicable to the artificial satellite too.

The earth is closer to the sun during winter. But heating effect is less because the sun's fall obliquely.

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:
\(F=\frac{G \times\left(\frac{4}{3} \pi \rho R^3\right) \times\left(\frac{32}{3} \pi \rho R^3\right)}{(3 R)^2} \)
\( \therefore F=\frac{128}{81} G \pi^2 R^4 \rho^2 \)

\({ T }_{ j }=11.6yr,{ r }_{ j }=?,{ T }_{ e }=1yr,{ r }_{ e }=1.5\times { 10 }^{ 11 }m\)
\(\\ \frac { { T }_{ j }^{ 2 } }{ { T }_{ e }^{ 2 } } =\frac { { r }_{ j }^{ 3 } }{ { r }_{ e }^{ 3 } } \)
\(\\ \Rightarrow \quad { r }_{ j }={ r }_{ e }\left( \frac { { { T }_{ j } } }{ { T }_{ e } } \right) ^{ 2/3 }=1.5\times 10^{ 11 }\times \left( \frac { 11.6 }{ 1 } \right) ^{ 2/3 }\)
\(\\ { r }_{ j }=7.68\times 10^{ 11 }m\) 

Height of the earth's satellite\(h=\left( \frac { { T }^{ 2 }R^{ 2 }g }{ 4\pi ^{ 2 } } \right) -R\)
Given,T= 90  min = \(90\times 60=5400s,R=6380km\)
\(g=9.8m/{ s }^{ 2 }=9.8\times { 10 }^{ -3 }km/{ s }^{ 2 }\)
Thus
 \(h=\left[ \frac { (5400)^{ 2 }\times (6380)^{ 2 }\times (9.8\times { 10 }^{ -3 } }{ 4\times 9.87 } \right] ^{ 1/3 }km-6380\quad km\)
or h = (6655 - 6380)km = 275 km

At all points inside a hollow spherical shell, potential is same. So, gravitational intensity, which is negative of gravitational potential gradient, is zero. Due to zero gravitational intensity, the gravitational forces acting on any particle at any point inside a spherical shell will be symmetrically placed. It follows from here that if we remove the upper hemispherical shell, the net gravitational force acting on a particle at P will be downwards. Since gravitational intensity is gravitational force per unit mass therefore, the direction of gravitational intensity will be along c. So, option (iii) is correct.

Potential energy of the comet changes as its kinetic energy changes.

The energy required to rocket an orbiting satellite out of gravitational influence is less than the energy required to project a stationary object,because incase of orbiting satelite,the gravitational pull of the earth acting on it is balanced by centripetal force,so work is required only in rocketing it(no work is required against the gravitational pull)

No, escape velocity is independent of the direction of projection

Step 1: Given
Mass of 1 st sphere, \(M_1=40 \mathrm{~kg}\)
Mass of 2nd sphere, \(M_2=15 \mathrm{~kg}\)
Gravitation force, \(F=\frac{1}{10} m g w t\)
Dis tan ce between centres of both spheres, R=0.2 m Universal gravitational cons \(\tan t, G=\) ?
Step 2: Formula used
Converting the value of force in SI unit (Newton),
\(F  =\frac{1}{10} \text { milligram wt } \)
\(=\frac{1}{10} \times 10^{-3} \mathrm{gwt} \)
\(=\frac{1}{10} \times 10^{-3} \times 10^{-3} \mathrm{kgwt}\)
\( =10^{-7} \times 9.8 \mathrm{~N} \)
\( =9.8 \times 10^{-7} \mathrm{~N}\)
Step 3: Calculations
Rearranging the Universal law of gravitation to find the value of G,
\(F =G \times \frac{M_1 \times M_2}{R^2} \)
\(G  =\frac{F \times R^2}{M_1 \times M_2}\)
Substituting the values in the expression,
\(G =\frac{F \times R^2}{M_1 \times M_2} \)
\(=\frac{9.8 \times 10^{-7} \mathrm{~N} \times(0.2 \mathrm{~m})^2}{40 \mathrm{~kg} \times 15 \mathrm{~kg}} \)
\(=6.533 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2}\)
Hence, the value of G is \(6.533 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2}\).