Explanation
The value of $${g_e}$$ on the surface of moon will be,
$$g' = g{\left( {1 + \frac{h}{R}} \right)^{ - 2}}$$
After putting the values of $$g,h,R$$
we get the value of $${g'}$$ as,
$${g'} = .0026\,{\rm{m}}/{{\rm{s}}^2}$$
This is almost negligible and can’t be felt on the person standing on moon.
Thus, a person on the surface of the moon does not feel the effect of earth's gravity because the gravity due to moon is much stronger.
A rocket is launched normal to the surface of the earth, away from the sun, along the line joining the sun and the earth. The sun is $$3 \times {10^5}$$ times heavier than the earth and is at a distance $$2.5 \times {10^4}$$ times larger than the radius of the earth. The escape velocity from earths gravitational field is $${{\rm{V}}_{\rm{c}}}\;{\rm{ = 11}}{\rm{.2km}}{{\rm{s}}^{{\rm{ - 1}}}}$$ . The minimum initial $$\left( {{{\rm{V}}_{\rm{c}}}} \right)\;$$ required for the rocket to be able to leave the sun-earth system is closest to (Ignore the rotation and revolution of the earth and the presence of any other planet)
Given that,
Tension T = 16.0 N
Now,
$$ T-2mg+2ma=0....(I) $$
$$ T-mg-ma=0....(II) $$
From equation (I) and (II)
$$ 3ma-mg=0 $$
$$ g=3a $$
$$ a=\dfrac{g}{3} $$
Now, put the value of g in equation (II)
$$ T-3ma-ma=0 $$
$$ T=4ma $$
$$ a=\dfrac{T}{4m} $$
Now, from equation of motion
At $$t= 1$$
$$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$
$$ s=0+\dfrac{1}{2}\times \dfrac{T}{4m}\times {{(1)}^{2}} $$
$$ s=\dfrac{16}{8m} $$
$$ s=\dfrac{2}{m} $$
Thus, change in the height of the block will be;
$$\Delta h=s$$
Net mass
$$2m-m=m$$
Now, decrease potential energy
$$ P.E=mg\Delta h $$
$$ P.E=m\times 9.8\times \dfrac{2}{m} $$
$$ P.E=19.6\,J $$
Hence, the decrease potential energy is $$19.6\ J$$
The acceleration due to gravity on the surface is given as
$$\dfrac{{Gm}}{{{r^2}}}$$
If the mass and the radius are doubled then the acceleration due to gravity will become
$$\dfrac{{G2m}}{{4{r^2}}}$$
Hence the acceleration becomes one half.
$${ \left( PE \right) }_{ height }={ \left( PE \right) }_{ surface }+{ \left( KE \right) }_{ surface }\\ -\dfrac { GMm }{ r } =-\dfrac { GMm }{ R } +\dfrac { 1 }{ 2 } m{ v }^{ 2 }\\ \therefore \dfrac { 1 }{ 2 } { v }^{ 2 }=GM\left( \dfrac { 1 }{ R } -\dfrac { 1 }{ r } \right) \\ \Rightarrow v=\sqrt { \dfrac { 2GM }{ R } \left( \dfrac { r-R }{ r } \right) } \dfrac { R }{ R } \\ \Rightarrow \sqrt { \dfrac { 2GMR }{ R } \left( \dfrac { r-R }{ r } \right) } \\ \Rightarrow \sqrt { \dfrac { 2gR(r-R) }{ r } } $$
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