**1. Why does one feel giddy while moving on a merry round?
**Ans – When moving in a merry go round, our weight appears to decrease when we move down and increase when we move up, this change in weight makes us feel giddy.

**2. The gravitational force between two bodies is 1N if the distance between them is doubled what will be the force between them?**

Ans – We know

F = GMm/r

^{2}

Cleary F∝ r

^{2}

∴ The force between two bodies becomes F/4 i.e. 1/4 N = 0.25 N

**3. If the density of a planet is doubled without any change in its radius, how does acceleration due to gravity (g) changes on the planet?**

Ans

We know g = GM/R^{2}

Mass = Volume x density

M = 4/3 π R^{3} x ρ

M = 4/3 π R^{3}ρ

g = G[4/3 π R^{3}ρ]/R^{2}

^{g = 4/3πRGρ
Clearly, On doubling density, the acceleration due to gravity (g) of the planet also get doubled. }

**4. The time period of the satellite is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, then what would be the new time period of the satellite. [Ans 40 hours]**

Ans – Using Kepler’s Third Law

T_{1}^{2} /T_{2}^{2} = R1^{3}/R2^{3}

Here T1 = 5 hr , R1= R

T2 = ? , R2= 4R

Putting these value

T1/T2 = (R1/R2)^{3/2}

5/T2 = (R/4R) ^{3/2}

T2/5 = (4)^{3/2}

T2 = 5 x (64)^{1/2
}T2 = 5 x 8

**T2 = 40 hours**

**5. If suddenly the gravitational force of attraction between earth and satellite becomes zero, what would happen to the satellite?**

Ans – The satellite will start moving in a straight line (with the same velocity) tangentially to the orbit where the force of attraction between earth and satellite becomes zero. OR The satellite will move tangentially to the original orbit with the same velocity.

**6. If the earth has a mass 9 times and radius 4 times than that of a planet “P”**.** Calculate the escape velocity at that planet “P”, if its value on earth 11.2 Km/s. [Ans 7.47 km/s]
**Ans

**We know, V**

_{e}= √2GM/R

V_{e} = 11.2 km/s

V_{p}=?

M_{P}=M_{e}/9

R_{P}=R_{e}/4

For Earth

V_{e}= √2GM_{e}/R_{e }– (1)

For Planet

V_{P} = √2GM_{P}/R_{P
} = √2G[M_{e}/9]/[R_{e}/4]

= [√2GM_{e}/R_{e}] x √4/9

V_{P} = 2V_{e}/3 [Using 1]

V_{P} = [11.2×2]/3

= 22.4/3

= 7.466 km/h

**V _{P }≈ 7.47 km/h**

**7. At what height from the surface of earth vill the value of ‘g’ reduced by 36 % of its value at the surface of the earth. [Ans 1600 km]
**Ans – Quite Easy. Do it by yourself.

**8. If the radius of earth shrinks by 2% mass remains constant. How would the value of acceleration due to gravity changes? Ans Increases by 4 %**

_{Ans – We know, g = GM/R2 Taking log both sides log g = log G + log M -2 log R Differentiating both sides dg/g = 0+0-2 dR/R dg/g = – 2dR/R dR/R = Decrease by 2% dR/R = -2/100}

Now

_{[dg x 100 %]/g = – 2[dR x 100%]/R
= -2 [-2/100] x 100 %
= 4 %}

Thus the value of g increases by 4%.

**9. How much faster than the present speed should the earth rotate so that a body lying on the equator may fly off into space. Ans 17 times the present velocity.**

~~We know ~~

~~g~~_{λ} = g – Rω^{2} cos λ

~~At equator, λ = 0~~^{°}

~~So g~~_{λ} = g – Rω^{2}

~~If g = 0 then we will feel weightlessness and will fly off from the euator.~~

~~So g~~_{λ }=0

~~g = Rω~~^{2}

~~or ω = √g/R~~

~~Will be Updated Soon.~~

Not in our current CBSE syllabus.