Q: At what distance above the surface of earth, the gravitational force will be reduced by \$10\%\$, if the radius of earth is \$6370\$ Km.

Gravitational force on the earth surface, \$F_{G_s}= \dfrac{GMm}{R^{2}}\$ Gravitational force at a height 'h' above, \$F_{G_{h}}= \dfrac{GMm}{(R+h)^{2}}\$ G.T. \$F_{G_{h}}= \dfrac{90}{100}(F_{G_{S}})\$ \$\Rightarrow \dfrac{GMm}{(R+h)^{2}}=\dfrac{9}{10}(\dfrac{GMm}{R^{2}})\$ \$\Rightarrow R+h=\sqrt{\dfrac{10}{9}}(R)\$ \$\Rightarrow 6370+h= \dfrac{3.162}{3} (6370)\$ \$\Rightarrow h=\dfrac{0.162 \times 6370}{3}=344\$ km

Q: The value of \$g\$ at a height \$h\$ above the surface of the earth is the same as at a depth \$d\$ below the surface of the earth. When both \$d\$ and \$h\$ are much smaller than the radius of earth, then which one of the following is correct

The value of gravity at a height \$h\$ above earth's surface is given by\$g_h=g\left(1-\dfrac {2h}{R}\right)\$and at a depth \$d\$ is given by \$g_d=g\left(1-\dfrac {d}{R}\right)\$Since these two values are same, we can see that \$d=2h\$

Q: A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to \$\sqrt{3}m\$ is.

Potential of a ring, \$V=\dfrac{-GM}{y}\$where, \$y=\sqrt{x^2+r^2}\$So initial, energy is \$\dfrac{-5G\times 1}{5}\$Final energy is \$\dfrac{-5G\times 1}{6}\$So the difference \$=\$ Final \$-\$ Initial \$=\dfrac{G}{6}\$

Q: The work done in shifting a particle of mass \$m\$ from the centre of earth to the surface of the earth is

Given,The mass of the particle be \$m\$,Initilly it is on the centre of the earth, soThe work done in moving a particle from center of earth to its surface\$\ W=\int_0^{R}Fdr\$\$=\int_0^R(-\dfrac{G(\rho\dfrac{4}{3}\pi r^3)m}{r^2})dr\$\$=\int_0^R(-\dfrac{4G\rho \pi mr}{3})dr\$\$=\dfrac{GMm}{2R},\$we know that \$F=\dfrac{GMm}{R^2}=mg\$\$=\dfrac{1}{2}mgR\$(Since \$g=\dfrac{GM}{R^2}\$)So, The work done in shifting a particle of mass from the centre of the earth to the surface of the earth is \$=\dfrac{mgR}{2}.\$

Q: The work done by an external agent to shift a point mass from infinity to the centre of the earth is W. Then choose the correct relation.

\$W_{ext}=\Delta U=U_{f}-U_{i}=U_{f}-0=U_{f}\$\$ W_{ext}=\displaystyle \frac{-3GMm}{2R} \rightarrow W_{ext}<0\$

Q: A spherical \$\Uparrow \mathrm{o}\mathrm{l}\mathrm{e}\$ of radius \$\mathrm{R}/2\$ is excavated from \$\mathrm{t}\Uparrow \mathrm{e}\$ asteroid of mass\$\mathrm{M}\$ as shown in fig. \$\mathrm{T}\Uparrow \mathrm{e}\$ gravitational acceleration at a point on \$\mathrm{t}\Uparrow \mathrm{e}\$surf ace of \$\mathrm{t}\Uparrow \mathrm{e}\$ asteroid just above \$\mathrm{t}\Uparrow \mathrm{e}\$ excavation is :

The energy at centre of hole is energy due to empty part \$\left( { \varepsilon }_{ 1 } \right) \$ plus energy due to remaining mass \$\left( { \varepsilon }_{ 2 } \right) \$.The energy due to empty mass \$=0\$ at center of the empty portion\$E={ E }_{ 1 }+{ E }_{ 2 }\$\${ E }_{ 2 }=E-{ E }_{ 1 }\$ \$=\dfrac { 1 }{ 8 } \dfrac { GM }{ { \left( \dfrac { R }{ 2 } \right) }^{ 2 } } \$\$E=\dfrac { GM }{ { 2R }^{ 2 } } \$

Q: If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is:

Acceleration due to gravity on earth's surface\$=g=\dfrac{GM}{R^2}\$.Gain in potential energy=\$\dfrac{GMm}{R}-\dfrac{GMm}{2R}=\dfrac{GMm}{2R}=\dfrac{1}{2}mgR\$

Q: A particle hanging from a massless spring stretches it by 2 cm at earths surface. How much will the same particle stretch the spring at a height of 2624 km from the surface of the earth? (Radius of earth \$=\$ 6400 km).

\$g= \dfrac{GM}{R^{2}}\$ and \$g_{h}=\displaystyle \dfrac{GM}{(R+h)^{2}}\$\$ \Rightarrow \dfrac{g_{h}}{g}=\left(\dfrac{R}{R+h}\right)^{2}=\left(\dfrac{6400}{6400+2624}\right)^{2}=\dfrac{1}{2}\$\$\Rightarrow g_{h}=\displaystyle \dfrac{g}{2}\$At the surface of earth, \$mg = kx\$At the height \$\mathrm{h}' \Rightarrow mg_{h}=kx^{'}\$\$\displaystyle \Rightarrow\dfrac{g_{h}}{g}=\frac{x^{'}}{x}\Rightarrow x^{'}=\left(\dfrac{g_{h}}{g}\right)x=\left(\frac{1}{2}\right)(2cm)=1cm\$

Q: Find the distance between the centre of gravity and centre of mass of a two-particle system attached to the ends of a light rod. Each particle has the same mass. Length of the rod is \$R\$, where \$R\$ is the radius of the earth.

The centre of gravity of the Earth lie at its centre (Point p in picture)The centre of mass of the rod is \$\cfrac { mR+m.0 }{ 2m } =\cfrac { R }{ 2 } \\ \$The distance between CM of the rod and CG of the Earth is=\$\cfrac { R }{ 2 } \$ [\$PS=\cfrac { R }{ 2 } \$ in the picture]

Question 1

At what distance above the surface of earth, the gravitational force will be reduced by $10\%$, if the radius of earth is $6370$ Km.

Accepted Answer

&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Gravitational force on&#10;the earth surface,&#10;&#10;$F_{G_s}=&#10;\dfrac{GMm}{R^{2}}$&#10;&#10;Gravitational force at&#10;a height 'h' above,&#10;&#10;$F_{G_{h}}=&#10;\dfrac{GMm}{(R+h)^{2}}$&#10;&#10;G.T. $F_{G_{h}}=&#10;\dfrac{90}{100}(F_{G_{S}})$&#10;&#10;$\Rightarrow&#10; \dfrac{GMm}{(R+h)^{2}}=\dfrac{9}{10}(\dfrac{GMm}{R^{2}})$&#10;&#10;$\Rightarrow&#10;R+h=\sqrt{\dfrac{10}{9}}(R)$&#10;&#10;$\Rightarrow 6370+h=&#10;\dfrac{3.162}{3} (6370)$&#10;&#10;$\Rightarrow&#10;h=\dfrac{0.162 \times 6370}{3}=344$  km&#10;&#10;

Question 2

The value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $d$ below the surface of the earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct

Accepted Answer

The value of gravity at a height $h$ above earth's surface is given by$g_h=g\left(1-\dfrac {2h}{R}\right)$and at a depth $d$ is given by $g_d=g\left(1-\dfrac {d}{R}\right)$Since these two values are same, we can see that $d=2h$

Question 3

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to $3 m$ is.

Accepted Answer

Potential of a ring, $V=\dfrac{-GM}{y}$where, $y=\sqrt{x^2+r^2}$So initial, energy is $\dfrac{-5G\times 1}{5}$Final energy is $\dfrac{-5G\times 1}{6}$So the difference $=$ Final $-$ Initial $=\dfrac{G}{6}$

Question 4

A particle of mass $10 gm$ is kept on the surface of a uniform sphere of mass $100 kg$ and radius $10 cm$. Find the work done against the gravitational force between them, to take the particle far away from the sphere. $\left ( G= 6.67\times10^{-11}Nm^{2}/Kg^{2} \right )$

Accepted Answer

At infinity, the gravitational potential energy of the system will be $0$.Initially, the particle is outside sphere and hence sphere will behave as point object at its centre,Therefore,${ E }_{ 0 }=-\dfrac { GMm }{ R } =-\dfrac { (6.67\times  { 10 }^{ -11 })\times(100)\times(0.01) }{ 0.1 } =-6.67\times{ 10 }^{ -10 }J$We know that, $W=\Delta U=U_{f}-U_{i} \implies W=-U_{i}$Thus, work done is $6.67\times { 10 }^{ -10 }J$

Question 5

The work done in shifting a particle of mass $m$ from the centre of earth to the surface of the earth is

Accepted Answer

Given,The mass of the particle be $m$,Initilly it is on the centre of the earth, soThe work done in moving a particle from center of earth to its surface$\ W=\int_0^{R}Fdr$$=\int_0^R(-\dfrac{G(\rho\dfrac{4}{3}\pi r^3)m}{r^2})dr$$=\int_0^R(-\dfrac{4G\rho \pi mr}{3})dr$$=\dfrac{GMm}{2R},$we know that $F=\dfrac{GMm}{R^2}=mg$$=\dfrac{1}{2}mgR$(Since $g=\dfrac{GM}{R^2}$)So, The work done in shifting a particle of mass from the centre of the earth to the surface of the earth is $=\dfrac{mgR}{2}.$

Question 6

The work done by an external agent to shift a point mass from infinity to the centre of the earth is W. Then choose the correct relation.

Accepted Answer

$W_{ext}=\Delta U=U_{f}-U_{i}=U_{f}-0=U_{f}$$ W_{ext}=\displaystyle \frac{-3GMm}{2R} \rightarrow W_{ext}<0$

Question 7

A spherical $\Uparrow \mathrm{o}\mathrm{l}\mathrm{e}$ of radius $\mathrm{R}/2$ is excavated from $\mathrm{t}\Uparrow \mathrm{e}$ asteroid of mass$\mathrm{M}$ as shown in fig. $\mathrm{T}\Uparrow \mathrm{e}$ gravitational acceleration at a point on $\mathrm{t}\Uparrow \mathrm{e}$surf ace of $\mathrm{t}\Uparrow \mathrm{e}$ asteroid just above $\mathrm{t}\Uparrow \mathrm{e}$ excavation is :

Accepted Answer

The energy at centre of hole is energy due to empty part $\left( { \varepsilon  }_{ 1 } \right) $ plus energy due to remaining mass $\left( { \varepsilon  }_{ 2 } \right) $.The energy due to empty mass $=0$ at center of the empty portion$E={ E }_{ 1 }+{ E }_{ 2 }$${ E }_{ 2 }=E-{ E }_{ 1 }$      $=\dfrac { 1 }{ 8 } \dfrac { GM }{ { \left( \dfrac { R }{ 2 }  \right)  }^{ 2 } } $$E=\dfrac { GM }{ { 2R }^{ 2 } } $

Question 8

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is:

Accepted Answer

Acceleration due to gravity on earth's surface$=g=\dfrac{GM}{R^2}$.Gain in potential energy=$\dfrac{GMm}{R}-\dfrac{GMm}{2R}=\dfrac{GMm}{2R}=\dfrac{1}{2}mgR$

Question 9

A particle hanging from a massless spring stretches it by 2 cm at earths surface. How much will the same particle stretch the spring at a height of 2624 km from the surface of the earth? (Radius of earth $=$ 6400 km).

Accepted Answer

$g= \dfrac{GM}{R^{2}}$ and $g_{h}=\displaystyle \dfrac{GM}{(R+h)^{2}}$$ \Rightarrow \dfrac{g_{h}}{g}=\left(\dfrac{R}{R+h}\right)^{2}=\left(\dfrac{6400}{6400+2624}\right)^{2}=\dfrac{1}{2}$$\Rightarrow g_{h}=\displaystyle \dfrac{g}{2}$At the surface of earth, $mg = kx$At the height $\mathrm{h}' \Rightarrow mg_{h}=kx^{'}$$\displaystyle \Rightarrow\dfrac{g_{h}}{g}=\frac{x^{'}}{x}\Rightarrow x^{'}=\left(\dfrac{g_{h}}{g}\right)x=\left(\frac{1}{2}\right)(2cm)=1cm$

Question 10

Find the distance between the centre of gravity and centre of mass of a two-particle system attached to the ends of a light rod. Each particle has the same mass. Length of the rod is $R$ , where $R$ is the radius of the earth.

Accepted Answer

The centre of gravity of the Earth lie at its centre (Point p in picture)The centre of mass of the rod is $\cfrac { mR+m.0 }{ 2m } =\cfrac { R }{ 2 } \ $The distance between CM of the rod and CG of the Earth is=$\cfrac { R }{ 2 } $ [$PS=\cfrac { R }{ 2 } $ in the picture]

Gravitation

Physics
2054 Views

At what distance above the surface of earth, the gravitational force will be reduced by $10 %$ , if the radius of earth is $6370$ Km.

The value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $d$ below the surface of the earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to $3 m$ is.

A particle of mass $10 g m$ is kept on the surface of a uniform sphere of mass $100 k g$ and radius $10 c m$ . Find the work done against the gravitational force between them, to take the particle far away from the sphere. $(G = 6.67 \times 1 0^{- 11} N m^{2} / K g^{2})$

The work done in shifting a particle of mass $m$ from the centre of earth to the surface of the earth is

The work done by an external agent to shift a point mass from infinity to the centre of the earth is W. Then choose the correct relation.

A spherical $⇑ o l e$ of radius $R / 2$ is excavated from $t ⇑ e$ asteroid of mass
$M$ as shown in fig. $T ⇑ e$ gravitational acceleration at a point on $t ⇑ e$
surf ace of $t ⇑ e$ asteroid just above $t ⇑ e$ excavation is :

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is:

A particle hanging from a massless spring stretches it by 2 cm at earths surface. How much will the same particle stretch the spring at a height of 2624 km from the surface of the earth? (Radius of earth $=$ 6400 km).

Find the distance between the centre of gravity and centre of mass of a two-particle system attached to the ends of a light rod. Each particle has the same mass. Length of the rod is $R$ , where $R$ is the radius of the earth.

Gravitation

Physics 2054 Views

At what distance above the surface of earth, the gravitational force will be reduced by 10%, if the radius of earth is 6370 Km.

The value of g at a height h above the surface of the earth is the same as at a depth d below the surface of the earth. When both d and h are much smaller than the radius of earth, then which one of the following is correct

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to 3​m is.

A particle of mass 10gm is kept on the surface of a uniform sphere of mass 100kg and radius 10cm. Find the work done against the gravitational force between them, to take the particle far away from the sphere. (G=6.67×10−11Nm2/Kg2)

The work done in shifting a particle of mass m from the centre of earth to the surface of the earth is

The work done by an external agent to shift a point mass from infinity to the centre of the earth is W. Then choose the correct relation.

A spherical ⇑ole of radius R/2 is excavated from t⇑e asteroid of mass M as shown in fig. T⇑e gravitational acceleration at a point on t⇑e surf ace of t⇑e asteroid just above t⇑e excavation is :

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth, is:

A particle hanging from a massless spring stretches it by 2 cm at earths surface. How much will the same particle stretch the spring at a height of 2624 km from the surface of the earth? (Radius of earth = 6400 km).

Find the distance between the centre of gravity and centre of mass of a two-particle system attached to the ends of a light rod. Each particle has the same mass. Length of the rod is R, where R is the radius of the earth.

Physics
2054 Views

At what distance above the surface of earth, the gravitational force will be reduced by $10 %$ , if the radius of earth is $6370$ Km.

The value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $d$ below the surface of the earth. When both $d$ and $h$ are much smaller than the radius of earth, then which one of the following is correct

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to $3 m$ is.

A particle of mass $10 g m$ is kept on the surface of a uniform sphere of mass $100 k g$ and radius $10 c m$ . Find the work done against the gravitational force between them, to take the particle far away from the sphere. $(G = 6.67 \times 1 0^{- 11} N m^{2} / K g^{2})$

The work done in shifting a particle of mass $m$ from the centre of earth to the surface of the earth is

A spherical $⇑ o l e$ of radius $R / 2$ is excavated from $t ⇑ e$ asteroid of mass
$M$ as shown in fig. $T ⇑ e$ gravitational acceleration at a point on $t ⇑ e$
surf ace of $t ⇑ e$ asteroid just above $t ⇑ e$ excavation is :

A particle hanging from a massless spring stretches it by 2 cm at earths surface. How much will the same particle stretch the spring at a height of 2624 km from the surface of the earth? (Radius of earth $=$ 6400 km).

Find the distance between the centre of gravity and centre of mass of a two-particle system attached to the ends of a light rod. Each particle has the same mass. Length of the rod is $R$ , where $R$ is the radius of the earth.