138368
For a planet, the acceleration due to gravity is half the acceleration due to gravity on the earth. Also the radius of the planet is half the radius of the earth. Then the mass of the planet in terms of mass of earth ' $M$ ' is
1 $\mathrm{M} / 7$
2 $\mathrm{M} / 6$
3 $M / 8$
4 $\mathrm{M} / 5$
Explanation:
C Given that, $\mathrm{g}_{\mathrm{p}}=\frac{1}{2} \mathrm{~g}_{\mathrm{e}}$ and $\mathrm{r}_{\mathrm{p}}=\frac{1}{2} \mathrm{r}_{\mathrm{e}}$ We know, $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ According to the question, $\frac{M_{p}}{M_{e}}=\frac{g_{p} r_{p}^{2}}{g_{e} r_{e}^{2}}$ $\frac{M_{p}}{M_{e}}=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}$ $M_{p}=\frac{M_{e}}{8}=\frac{M}{8}$
MHT-CET 2019
Gravitation
138369
The masses of two planets are in the ratio 1:2. Their radii are in the ratio $1: 2$. The acceleration due to gravity on the planets are in the ratio.
1 $1: 2$
2 $2: 1$
3 $3: 5$
4 $5: 3$
Explanation:
B The expression of ' $\mathrm{g}$ ' on the Earth is, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Let us Consider $M_{1}$ and $M_{2}$ be the masses and $R_{1}$ and $R_{2}$ be the radii of the planets respectively. So, $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}^{2}}$ And $\quad \mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{R}_{2}^{2}}$ Given that, $\mathrm{M}_{2}=2 \mathrm{M}_{1} \quad$ and $\quad \mathrm{R}_{2}=2 \mathrm{R}_{1}$ Substituting these in the equation (ii), $\mathrm{g}_{2}=\frac{\mathrm{G} \times 2 \mathrm{M}_{1}}{\left(2 \mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \mathrm{~g}_{1} \quad\left[\text { from eq }{ }^{\mathrm{n}}(\mathrm{i}) \mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}\right]$ $\mathrm{g}_{1}: \mathrm{g}_{2}=2: 1$
MHT-CET 2014
Gravitation
138370
Calculate angular velocity of earth so that acceleration due to gravity at $60^{\circ}$ latitude becomes zero. (Radius of earth $=6400 \mathrm{~km}$, gravitational acceleration at poles $=10 \mathrm{~m} / \mathrm{s}^{2}$, $\cos 60^{\circ}=0.5$ )
1 $7.8 \times 10^{-2} \mathrm{rad} / \mathrm{s}$
2 $0.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
3 $1 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
4 $2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
Explanation:
D Given that, $\mathrm{g}^{\prime}=0 \text {, at } \lambda=60^{\circ}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We, know that new acceleration due to gravity $g^{\prime}$ is given- $g^{\prime}=g-R_{e} \omega^{2} \cos ^{2} \lambda$ $\therefore \quad 0 =g-R_{e} \omega^{2} \cos ^{2} 60$ $\omega^{2} =\frac{g}{R_{e} \times \cos ^{2} 60^{\circ}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times\left(\frac{1}{2}\right)^{2}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times 0.5 \times 0.5}$ $\omega =\sqrt{\frac{10}{6.4 \times 10^{6} \times 0.25}}$ $\omega =2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
MHT-CET 2014
Gravitation
138371
If $g$ is the acceleration due to gravity on earth's surface, the gain of 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
1 $2 \mathrm{mgR}$
2 $\mathrm{mgR}$
3 $\frac{1}{2} \mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
C Initial potential energy on earth surface $\mathrm{U}_{\mathrm{i}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ Final potential energy at height $\mathrm{h}$ from earth surface $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}} \quad(\mathrm{h}=\mathrm{R})$ $\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gR}^{2} \mathrm{~m}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gmR}}{2}$ ${\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]}$
138368
For a planet, the acceleration due to gravity is half the acceleration due to gravity on the earth. Also the radius of the planet is half the radius of the earth. Then the mass of the planet in terms of mass of earth ' $M$ ' is
1 $\mathrm{M} / 7$
2 $\mathrm{M} / 6$
3 $M / 8$
4 $\mathrm{M} / 5$
Explanation:
C Given that, $\mathrm{g}_{\mathrm{p}}=\frac{1}{2} \mathrm{~g}_{\mathrm{e}}$ and $\mathrm{r}_{\mathrm{p}}=\frac{1}{2} \mathrm{r}_{\mathrm{e}}$ We know, $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ According to the question, $\frac{M_{p}}{M_{e}}=\frac{g_{p} r_{p}^{2}}{g_{e} r_{e}^{2}}$ $\frac{M_{p}}{M_{e}}=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}$ $M_{p}=\frac{M_{e}}{8}=\frac{M}{8}$
MHT-CET 2019
Gravitation
138369
The masses of two planets are in the ratio 1:2. Their radii are in the ratio $1: 2$. The acceleration due to gravity on the planets are in the ratio.
1 $1: 2$
2 $2: 1$
3 $3: 5$
4 $5: 3$
Explanation:
B The expression of ' $\mathrm{g}$ ' on the Earth is, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Let us Consider $M_{1}$ and $M_{2}$ be the masses and $R_{1}$ and $R_{2}$ be the radii of the planets respectively. So, $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}^{2}}$ And $\quad \mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{R}_{2}^{2}}$ Given that, $\mathrm{M}_{2}=2 \mathrm{M}_{1} \quad$ and $\quad \mathrm{R}_{2}=2 \mathrm{R}_{1}$ Substituting these in the equation (ii), $\mathrm{g}_{2}=\frac{\mathrm{G} \times 2 \mathrm{M}_{1}}{\left(2 \mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \mathrm{~g}_{1} \quad\left[\text { from eq }{ }^{\mathrm{n}}(\mathrm{i}) \mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}\right]$ $\mathrm{g}_{1}: \mathrm{g}_{2}=2: 1$
MHT-CET 2014
Gravitation
138370
Calculate angular velocity of earth so that acceleration due to gravity at $60^{\circ}$ latitude becomes zero. (Radius of earth $=6400 \mathrm{~km}$, gravitational acceleration at poles $=10 \mathrm{~m} / \mathrm{s}^{2}$, $\cos 60^{\circ}=0.5$ )
1 $7.8 \times 10^{-2} \mathrm{rad} / \mathrm{s}$
2 $0.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
3 $1 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
4 $2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
Explanation:
D Given that, $\mathrm{g}^{\prime}=0 \text {, at } \lambda=60^{\circ}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We, know that new acceleration due to gravity $g^{\prime}$ is given- $g^{\prime}=g-R_{e} \omega^{2} \cos ^{2} \lambda$ $\therefore \quad 0 =g-R_{e} \omega^{2} \cos ^{2} 60$ $\omega^{2} =\frac{g}{R_{e} \times \cos ^{2} 60^{\circ}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times\left(\frac{1}{2}\right)^{2}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times 0.5 \times 0.5}$ $\omega =\sqrt{\frac{10}{6.4 \times 10^{6} \times 0.25}}$ $\omega =2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
MHT-CET 2014
Gravitation
138371
If $g$ is the acceleration due to gravity on earth's surface, the gain of 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
1 $2 \mathrm{mgR}$
2 $\mathrm{mgR}$
3 $\frac{1}{2} \mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
C Initial potential energy on earth surface $\mathrm{U}_{\mathrm{i}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ Final potential energy at height $\mathrm{h}$ from earth surface $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}} \quad(\mathrm{h}=\mathrm{R})$ $\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gR}^{2} \mathrm{~m}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gmR}}{2}$ ${\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]}$
138368
For a planet, the acceleration due to gravity is half the acceleration due to gravity on the earth. Also the radius of the planet is half the radius of the earth. Then the mass of the planet in terms of mass of earth ' $M$ ' is
1 $\mathrm{M} / 7$
2 $\mathrm{M} / 6$
3 $M / 8$
4 $\mathrm{M} / 5$
Explanation:
C Given that, $\mathrm{g}_{\mathrm{p}}=\frac{1}{2} \mathrm{~g}_{\mathrm{e}}$ and $\mathrm{r}_{\mathrm{p}}=\frac{1}{2} \mathrm{r}_{\mathrm{e}}$ We know, $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ According to the question, $\frac{M_{p}}{M_{e}}=\frac{g_{p} r_{p}^{2}}{g_{e} r_{e}^{2}}$ $\frac{M_{p}}{M_{e}}=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}$ $M_{p}=\frac{M_{e}}{8}=\frac{M}{8}$
MHT-CET 2019
Gravitation
138369
The masses of two planets are in the ratio 1:2. Their radii are in the ratio $1: 2$. The acceleration due to gravity on the planets are in the ratio.
1 $1: 2$
2 $2: 1$
3 $3: 5$
4 $5: 3$
Explanation:
B The expression of ' $\mathrm{g}$ ' on the Earth is, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Let us Consider $M_{1}$ and $M_{2}$ be the masses and $R_{1}$ and $R_{2}$ be the radii of the planets respectively. So, $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}^{2}}$ And $\quad \mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{R}_{2}^{2}}$ Given that, $\mathrm{M}_{2}=2 \mathrm{M}_{1} \quad$ and $\quad \mathrm{R}_{2}=2 \mathrm{R}_{1}$ Substituting these in the equation (ii), $\mathrm{g}_{2}=\frac{\mathrm{G} \times 2 \mathrm{M}_{1}}{\left(2 \mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \mathrm{~g}_{1} \quad\left[\text { from eq }{ }^{\mathrm{n}}(\mathrm{i}) \mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}\right]$ $\mathrm{g}_{1}: \mathrm{g}_{2}=2: 1$
MHT-CET 2014
Gravitation
138370
Calculate angular velocity of earth so that acceleration due to gravity at $60^{\circ}$ latitude becomes zero. (Radius of earth $=6400 \mathrm{~km}$, gravitational acceleration at poles $=10 \mathrm{~m} / \mathrm{s}^{2}$, $\cos 60^{\circ}=0.5$ )
1 $7.8 \times 10^{-2} \mathrm{rad} / \mathrm{s}$
2 $0.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
3 $1 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
4 $2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
Explanation:
D Given that, $\mathrm{g}^{\prime}=0 \text {, at } \lambda=60^{\circ}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We, know that new acceleration due to gravity $g^{\prime}$ is given- $g^{\prime}=g-R_{e} \omega^{2} \cos ^{2} \lambda$ $\therefore \quad 0 =g-R_{e} \omega^{2} \cos ^{2} 60$ $\omega^{2} =\frac{g}{R_{e} \times \cos ^{2} 60^{\circ}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times\left(\frac{1}{2}\right)^{2}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times 0.5 \times 0.5}$ $\omega =\sqrt{\frac{10}{6.4 \times 10^{6} \times 0.25}}$ $\omega =2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
MHT-CET 2014
Gravitation
138371
If $g$ is the acceleration due to gravity on earth's surface, the gain of 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
1 $2 \mathrm{mgR}$
2 $\mathrm{mgR}$
3 $\frac{1}{2} \mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
C Initial potential energy on earth surface $\mathrm{U}_{\mathrm{i}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ Final potential energy at height $\mathrm{h}$ from earth surface $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}} \quad(\mathrm{h}=\mathrm{R})$ $\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gR}^{2} \mathrm{~m}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gmR}}{2}$ ${\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]}$
138368
For a planet, the acceleration due to gravity is half the acceleration due to gravity on the earth. Also the radius of the planet is half the radius of the earth. Then the mass of the planet in terms of mass of earth ' $M$ ' is
1 $\mathrm{M} / 7$
2 $\mathrm{M} / 6$
3 $M / 8$
4 $\mathrm{M} / 5$
Explanation:
C Given that, $\mathrm{g}_{\mathrm{p}}=\frac{1}{2} \mathrm{~g}_{\mathrm{e}}$ and $\mathrm{r}_{\mathrm{p}}=\frac{1}{2} \mathrm{r}_{\mathrm{e}}$ We know, $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ According to the question, $\frac{M_{p}}{M_{e}}=\frac{g_{p} r_{p}^{2}}{g_{e} r_{e}^{2}}$ $\frac{M_{p}}{M_{e}}=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}$ $M_{p}=\frac{M_{e}}{8}=\frac{M}{8}$
MHT-CET 2019
Gravitation
138369
The masses of two planets are in the ratio 1:2. Their radii are in the ratio $1: 2$. The acceleration due to gravity on the planets are in the ratio.
1 $1: 2$
2 $2: 1$
3 $3: 5$
4 $5: 3$
Explanation:
B The expression of ' $\mathrm{g}$ ' on the Earth is, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Let us Consider $M_{1}$ and $M_{2}$ be the masses and $R_{1}$ and $R_{2}$ be the radii of the planets respectively. So, $\mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}^{2}}$ And $\quad \mathrm{g}_{2}=\frac{\mathrm{GM}_{2}}{\mathrm{R}_{2}^{2}}$ Given that, $\mathrm{M}_{2}=2 \mathrm{M}_{1} \quad$ and $\quad \mathrm{R}_{2}=2 \mathrm{R}_{1}$ Substituting these in the equation (ii), $\mathrm{g}_{2}=\frac{\mathrm{G} \times 2 \mathrm{M}_{1}}{\left(2 \mathrm{R}_{1}\right)^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}$ $\mathrm{~g}_{2}=\frac{1}{2} \mathrm{~g}_{1} \quad\left[\text { from eq }{ }^{\mathrm{n}}(\mathrm{i}) \mathrm{g}_{1}=\frac{\mathrm{GM}_{1}}{\mathrm{R}_{1}{ }^{2}}\right]$ $\mathrm{g}_{1}: \mathrm{g}_{2}=2: 1$
MHT-CET 2014
Gravitation
138370
Calculate angular velocity of earth so that acceleration due to gravity at $60^{\circ}$ latitude becomes zero. (Radius of earth $=6400 \mathrm{~km}$, gravitational acceleration at poles $=10 \mathrm{~m} / \mathrm{s}^{2}$, $\cos 60^{\circ}=0.5$ )
1 $7.8 \times 10^{-2} \mathrm{rad} / \mathrm{s}$
2 $0.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
3 $1 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
4 $2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
Explanation:
D Given that, $\mathrm{g}^{\prime}=0 \text {, at } \lambda=60^{\circ}$ Radius of earth $\left(\mathrm{R}_{\mathrm{e}}\right)=6400 \mathrm{~km}$ We, know that new acceleration due to gravity $g^{\prime}$ is given- $g^{\prime}=g-R_{e} \omega^{2} \cos ^{2} \lambda$ $\therefore \quad 0 =g-R_{e} \omega^{2} \cos ^{2} 60$ $\omega^{2} =\frac{g}{R_{e} \times \cos ^{2} 60^{\circ}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times\left(\frac{1}{2}\right)^{2}}$ $\omega^{2} =\frac{10}{6.4 \times 10^{6} \times 0.5 \times 0.5}$ $\omega =\sqrt{\frac{10}{6.4 \times 10^{6} \times 0.25}}$ $\omega =2.5 \times 10^{-3} \mathrm{rad} / \mathrm{s}$
MHT-CET 2014
Gravitation
138371
If $g$ is the acceleration due to gravity on earth's surface, the gain of 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
1 $2 \mathrm{mgR}$
2 $\mathrm{mgR}$
3 $\frac{1}{2} \mathrm{mgR}$
4 $\frac{1}{4} \mathrm{mgR}$
Explanation:
C Initial potential energy on earth surface $\mathrm{U}_{\mathrm{i}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ Final potential energy at height $\mathrm{h}$ from earth surface $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}$ $\mathrm{U}_{\mathrm{f}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}} \quad(\mathrm{h}=\mathrm{R})$ $\Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)$ $\Delta \mathrm{U}=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gR}^{2} \mathrm{~m}}{2 \mathrm{R}}$ $\Delta \mathrm{U}=\frac{\mathrm{gmR}}{2}$ ${\left[\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right]}$
