138387
A man weight $W$ on the surface of earth. His weight at a height equal to $R$ (earth's radius) is
1 $\mathrm{W}$
2 $\mathrm{Wt}^{2}$
3 $\mathrm{W} / 4$
4 $\mathrm{W} / 8$
Explanation:
C $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g^{\prime}}{g}=\left(\frac{1}{1+\frac{h}{R}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(\frac{1}{1+\frac{\mathrm{R}}{\mathrm{R}}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{4}$ $[\because \mathrm{h}=\mathrm{R}]$ Multiply by mass (m) $\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{4}$ $\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{1}{4}$ $\mathrm{~W}^{\prime}=\frac{\mathrm{W}}{4}$
Assam CEE-2019
Gravitation
138389
There is a planet which is 8 times massive and 27 times denser than the earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surface of the planet and the earth respectively then
1 $g^{\prime}=8 \mathrm{~g}$
2 $\mathrm{g}^{\prime}=27 \mathrm{~g}$
3 $\mathrm{g}^{\prime}=18 \mathrm{~g}$
4 $\mathrm{g}^{\prime}=\frac{9}{4} \mathrm{~g}$
Explanation:
C Given, Mass of planet $\left(\mathrm{m}_{\mathrm{p}}\right)=8 \mathrm{~m}_{\mathrm{e}}$ Density of planet $\left(D_{p}\right)=27 D_{e}$ $\frac{\mathrm{m}_{\mathrm{p}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{8 \mathrm{~m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{3}} {\left[\because \mathrm{m}_{\mathrm{p}}=8 \mathrm{~m}_{\mathrm{e}}\right]}$ $r_{e}^{3}=\frac{27}{8} r_{p}^{3}$ $r_{e}=\frac{3}{2} r_{p}$ $\because \quad \mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{r}^{2}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{Gm}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{p}}^{2}}}{\frac{\mathrm{Gm} \mathrm{e}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{2}}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{e}}^2}{\mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{p}}^2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \times\left(\frac{3}{2}\right)^2 \mathrm{r}_{\mathrm{p}}^2}{\mathrm{r}_{\mathrm{p}}^2} \quad\left[\because \mathrm{r}_{\mathrm{e}}=\frac{3}{2} \mathrm{r}_{\mathrm{p}}\right]$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=18$ $\mathrm{~g}^{\prime}=18 \mathrm{~g}$
AP EAMCET-07.07.2022
Gravitation
138390
The mass and diameter of a planet $P$ are twice and thrice to that of the corresponding parameters of earth respectively. If the acceleration due to gravity on earth's surface is $10 \mathrm{~ms}^{-2}$, the acceleration due to gravity on the surface of the planet $P$ is
1 $4.81 \mathrm{~m} \mathrm{~s}^{-2}$
2 $9.81 \mathrm{~m} \mathrm{~s}$
3 $2.22 \mathrm{~m} \mathrm{~s}^{-2}$
4 $6.32 \mathrm{~m} \mathrm{~s}^{-2}$
Explanation:
C Given that $\mathrm{g}_{\mathrm{e}}=10 \mathrm{~m} / \mathrm{sec}^{2}$ Mass of earth $=M_{e}$ Radius of the earth $=R_{e}$ We know that $\mathrm{g}_{\mathrm{e}}=$ gravity of the earth can be given by- $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ For planet $\mathrm{P}$, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{G} \mathrm{M}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ According to question, $\mathrm{M}_{\mathrm{P}}=2 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{P}}=3 \mathrm{R}_{\mathrm{e}}$ Now, $\quad g_{P}=\frac{G \times 2 M_{e}}{\left(3 R_{e}\right)^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}^{2}}$ Divide equation (i) by equation (ii), we get- $\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{P}}}=\frac{9}{2}$ $\frac{10}{\mathrm{~g}_{\mathrm{P}}}=\frac{9}{2} \Rightarrow \mathrm{g}_{\mathrm{P}}=2.22 \mathrm{~m} / \mathrm{sec}^{2}$
AP EAMCET-07.09.2021
Gravitation
138391
At a distance $320 \mathrm{~km}$ above the surface of the earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly - (earth's radius $=6400$ km)
1 $2 \%$
2 $6 \%$
3 $10 \%$
4 $14 \%$
Explanation:
C Height above the surface $(\mathrm{h})=320 \mathrm{~km}$ As we know for smallest height, $\frac{g^{\prime}}{g}=\left(1-\frac{2 h}{R}\right)=\frac{9}{10}$ $g^{\prime}=\frac{9}{10} g$ $\therefore \% \text { decrease in } g=\frac{g-g^{\prime}}{g} \times 100$ $=\frac{g-\frac{9}{10} g}{g} \times 100$ $=10 \%$
138387
A man weight $W$ on the surface of earth. His weight at a height equal to $R$ (earth's radius) is
1 $\mathrm{W}$
2 $\mathrm{Wt}^{2}$
3 $\mathrm{W} / 4$
4 $\mathrm{W} / 8$
Explanation:
C $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g^{\prime}}{g}=\left(\frac{1}{1+\frac{h}{R}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(\frac{1}{1+\frac{\mathrm{R}}{\mathrm{R}}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{4}$ $[\because \mathrm{h}=\mathrm{R}]$ Multiply by mass (m) $\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{4}$ $\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{1}{4}$ $\mathrm{~W}^{\prime}=\frac{\mathrm{W}}{4}$
Assam CEE-2019
Gravitation
138389
There is a planet which is 8 times massive and 27 times denser than the earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surface of the planet and the earth respectively then
1 $g^{\prime}=8 \mathrm{~g}$
2 $\mathrm{g}^{\prime}=27 \mathrm{~g}$
3 $\mathrm{g}^{\prime}=18 \mathrm{~g}$
4 $\mathrm{g}^{\prime}=\frac{9}{4} \mathrm{~g}$
Explanation:
C Given, Mass of planet $\left(\mathrm{m}_{\mathrm{p}}\right)=8 \mathrm{~m}_{\mathrm{e}}$ Density of planet $\left(D_{p}\right)=27 D_{e}$ $\frac{\mathrm{m}_{\mathrm{p}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{8 \mathrm{~m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{3}} {\left[\because \mathrm{m}_{\mathrm{p}}=8 \mathrm{~m}_{\mathrm{e}}\right]}$ $r_{e}^{3}=\frac{27}{8} r_{p}^{3}$ $r_{e}=\frac{3}{2} r_{p}$ $\because \quad \mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{r}^{2}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{Gm}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{p}}^{2}}}{\frac{\mathrm{Gm} \mathrm{e}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{2}}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{e}}^2}{\mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{p}}^2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \times\left(\frac{3}{2}\right)^2 \mathrm{r}_{\mathrm{p}}^2}{\mathrm{r}_{\mathrm{p}}^2} \quad\left[\because \mathrm{r}_{\mathrm{e}}=\frac{3}{2} \mathrm{r}_{\mathrm{p}}\right]$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=18$ $\mathrm{~g}^{\prime}=18 \mathrm{~g}$
AP EAMCET-07.07.2022
Gravitation
138390
The mass and diameter of a planet $P$ are twice and thrice to that of the corresponding parameters of earth respectively. If the acceleration due to gravity on earth's surface is $10 \mathrm{~ms}^{-2}$, the acceleration due to gravity on the surface of the planet $P$ is
1 $4.81 \mathrm{~m} \mathrm{~s}^{-2}$
2 $9.81 \mathrm{~m} \mathrm{~s}$
3 $2.22 \mathrm{~m} \mathrm{~s}^{-2}$
4 $6.32 \mathrm{~m} \mathrm{~s}^{-2}$
Explanation:
C Given that $\mathrm{g}_{\mathrm{e}}=10 \mathrm{~m} / \mathrm{sec}^{2}$ Mass of earth $=M_{e}$ Radius of the earth $=R_{e}$ We know that $\mathrm{g}_{\mathrm{e}}=$ gravity of the earth can be given by- $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ For planet $\mathrm{P}$, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{G} \mathrm{M}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ According to question, $\mathrm{M}_{\mathrm{P}}=2 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{P}}=3 \mathrm{R}_{\mathrm{e}}$ Now, $\quad g_{P}=\frac{G \times 2 M_{e}}{\left(3 R_{e}\right)^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}^{2}}$ Divide equation (i) by equation (ii), we get- $\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{P}}}=\frac{9}{2}$ $\frac{10}{\mathrm{~g}_{\mathrm{P}}}=\frac{9}{2} \Rightarrow \mathrm{g}_{\mathrm{P}}=2.22 \mathrm{~m} / \mathrm{sec}^{2}$
AP EAMCET-07.09.2021
Gravitation
138391
At a distance $320 \mathrm{~km}$ above the surface of the earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly - (earth's radius $=6400$ km)
1 $2 \%$
2 $6 \%$
3 $10 \%$
4 $14 \%$
Explanation:
C Height above the surface $(\mathrm{h})=320 \mathrm{~km}$ As we know for smallest height, $\frac{g^{\prime}}{g}=\left(1-\frac{2 h}{R}\right)=\frac{9}{10}$ $g^{\prime}=\frac{9}{10} g$ $\therefore \% \text { decrease in } g=\frac{g-g^{\prime}}{g} \times 100$ $=\frac{g-\frac{9}{10} g}{g} \times 100$ $=10 \%$
138387
A man weight $W$ on the surface of earth. His weight at a height equal to $R$ (earth's radius) is
1 $\mathrm{W}$
2 $\mathrm{Wt}^{2}$
3 $\mathrm{W} / 4$
4 $\mathrm{W} / 8$
Explanation:
C $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g^{\prime}}{g}=\left(\frac{1}{1+\frac{h}{R}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(\frac{1}{1+\frac{\mathrm{R}}{\mathrm{R}}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{4}$ $[\because \mathrm{h}=\mathrm{R}]$ Multiply by mass (m) $\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{4}$ $\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{1}{4}$ $\mathrm{~W}^{\prime}=\frac{\mathrm{W}}{4}$
Assam CEE-2019
Gravitation
138389
There is a planet which is 8 times massive and 27 times denser than the earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surface of the planet and the earth respectively then
1 $g^{\prime}=8 \mathrm{~g}$
2 $\mathrm{g}^{\prime}=27 \mathrm{~g}$
3 $\mathrm{g}^{\prime}=18 \mathrm{~g}$
4 $\mathrm{g}^{\prime}=\frac{9}{4} \mathrm{~g}$
Explanation:
C Given, Mass of planet $\left(\mathrm{m}_{\mathrm{p}}\right)=8 \mathrm{~m}_{\mathrm{e}}$ Density of planet $\left(D_{p}\right)=27 D_{e}$ $\frac{\mathrm{m}_{\mathrm{p}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{8 \mathrm{~m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{3}} {\left[\because \mathrm{m}_{\mathrm{p}}=8 \mathrm{~m}_{\mathrm{e}}\right]}$ $r_{e}^{3}=\frac{27}{8} r_{p}^{3}$ $r_{e}=\frac{3}{2} r_{p}$ $\because \quad \mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{r}^{2}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{Gm}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{p}}^{2}}}{\frac{\mathrm{Gm} \mathrm{e}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{2}}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{e}}^2}{\mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{p}}^2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \times\left(\frac{3}{2}\right)^2 \mathrm{r}_{\mathrm{p}}^2}{\mathrm{r}_{\mathrm{p}}^2} \quad\left[\because \mathrm{r}_{\mathrm{e}}=\frac{3}{2} \mathrm{r}_{\mathrm{p}}\right]$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=18$ $\mathrm{~g}^{\prime}=18 \mathrm{~g}$
AP EAMCET-07.07.2022
Gravitation
138390
The mass and diameter of a planet $P$ are twice and thrice to that of the corresponding parameters of earth respectively. If the acceleration due to gravity on earth's surface is $10 \mathrm{~ms}^{-2}$, the acceleration due to gravity on the surface of the planet $P$ is
1 $4.81 \mathrm{~m} \mathrm{~s}^{-2}$
2 $9.81 \mathrm{~m} \mathrm{~s}$
3 $2.22 \mathrm{~m} \mathrm{~s}^{-2}$
4 $6.32 \mathrm{~m} \mathrm{~s}^{-2}$
Explanation:
C Given that $\mathrm{g}_{\mathrm{e}}=10 \mathrm{~m} / \mathrm{sec}^{2}$ Mass of earth $=M_{e}$ Radius of the earth $=R_{e}$ We know that $\mathrm{g}_{\mathrm{e}}=$ gravity of the earth can be given by- $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ For planet $\mathrm{P}$, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{G} \mathrm{M}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ According to question, $\mathrm{M}_{\mathrm{P}}=2 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{P}}=3 \mathrm{R}_{\mathrm{e}}$ Now, $\quad g_{P}=\frac{G \times 2 M_{e}}{\left(3 R_{e}\right)^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}^{2}}$ Divide equation (i) by equation (ii), we get- $\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{P}}}=\frac{9}{2}$ $\frac{10}{\mathrm{~g}_{\mathrm{P}}}=\frac{9}{2} \Rightarrow \mathrm{g}_{\mathrm{P}}=2.22 \mathrm{~m} / \mathrm{sec}^{2}$
AP EAMCET-07.09.2021
Gravitation
138391
At a distance $320 \mathrm{~km}$ above the surface of the earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly - (earth's radius $=6400$ km)
1 $2 \%$
2 $6 \%$
3 $10 \%$
4 $14 \%$
Explanation:
C Height above the surface $(\mathrm{h})=320 \mathrm{~km}$ As we know for smallest height, $\frac{g^{\prime}}{g}=\left(1-\frac{2 h}{R}\right)=\frac{9}{10}$ $g^{\prime}=\frac{9}{10} g$ $\therefore \% \text { decrease in } g=\frac{g-g^{\prime}}{g} \times 100$ $=\frac{g-\frac{9}{10} g}{g} \times 100$ $=10 \%$
138387
A man weight $W$ on the surface of earth. His weight at a height equal to $R$ (earth's radius) is
1 $\mathrm{W}$
2 $\mathrm{Wt}^{2}$
3 $\mathrm{W} / 4$
4 $\mathrm{W} / 8$
Explanation:
C $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g^{\prime}}{g}=\left(\frac{1}{1+\frac{h}{R}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\left(\frac{1}{1+\frac{\mathrm{R}}{\mathrm{R}}}\right)^{2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{1}{4}$ $[\because \mathrm{h}=\mathrm{R}]$ Multiply by mass (m) $\frac{\mathrm{mg}^{\prime}}{\mathrm{mg}}=\frac{1}{4}$ $\frac{\mathrm{W}^{\prime}}{\mathrm{W}}=\frac{1}{4}$ $\mathrm{~W}^{\prime}=\frac{\mathrm{W}}{4}$
Assam CEE-2019
Gravitation
138389
There is a planet which is 8 times massive and 27 times denser than the earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surface of the planet and the earth respectively then
1 $g^{\prime}=8 \mathrm{~g}$
2 $\mathrm{g}^{\prime}=27 \mathrm{~g}$
3 $\mathrm{g}^{\prime}=18 \mathrm{~g}$
4 $\mathrm{g}^{\prime}=\frac{9}{4} \mathrm{~g}$
Explanation:
C Given, Mass of planet $\left(\mathrm{m}_{\mathrm{p}}\right)=8 \mathrm{~m}_{\mathrm{e}}$ Density of planet $\left(D_{p}\right)=27 D_{e}$ $\frac{\mathrm{m}_{\mathrm{p}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\frac{4}{3} \pi \mathrm{r}_{\mathrm{e}}^{3}}$ $\frac{8 \mathrm{~m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{p}}^{3}}=27 \frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{3}} {\left[\because \mathrm{m}_{\mathrm{p}}=8 \mathrm{~m}_{\mathrm{e}}\right]}$ $r_{e}^{3}=\frac{27}{8} r_{p}^{3}$ $r_{e}=\frac{3}{2} r_{p}$ $\because \quad \mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{r}^{2}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\frac{\mathrm{Gm}_{\mathrm{p}}}{\mathrm{r}_{\mathrm{p}}^{2}}}{\frac{\mathrm{Gm} \mathrm{e}_{\mathrm{e}}}{\mathrm{r}_{\mathrm{e}}^{2}}}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{e}}^2}{\mathrm{~m}_{\mathrm{e}} \times \mathrm{r}_{\mathrm{p}}^2}$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{8 \times\left(\frac{3}{2}\right)^2 \mathrm{r}_{\mathrm{p}}^2}{\mathrm{r}_{\mathrm{p}}^2} \quad\left[\because \mathrm{r}_{\mathrm{e}}=\frac{3}{2} \mathrm{r}_{\mathrm{p}}\right]$ $\frac{\mathrm{g}^{\prime}}{\mathrm{g}}=18$ $\mathrm{~g}^{\prime}=18 \mathrm{~g}$
AP EAMCET-07.07.2022
Gravitation
138390
The mass and diameter of a planet $P$ are twice and thrice to that of the corresponding parameters of earth respectively. If the acceleration due to gravity on earth's surface is $10 \mathrm{~ms}^{-2}$, the acceleration due to gravity on the surface of the planet $P$ is
1 $4.81 \mathrm{~m} \mathrm{~s}^{-2}$
2 $9.81 \mathrm{~m} \mathrm{~s}$
3 $2.22 \mathrm{~m} \mathrm{~s}^{-2}$
4 $6.32 \mathrm{~m} \mathrm{~s}^{-2}$
Explanation:
C Given that $\mathrm{g}_{\mathrm{e}}=10 \mathrm{~m} / \mathrm{sec}^{2}$ Mass of earth $=M_{e}$ Radius of the earth $=R_{e}$ We know that $\mathrm{g}_{\mathrm{e}}=$ gravity of the earth can be given by- $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}{ }^{2}}$ For planet $\mathrm{P}$, $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{G} \mathrm{M}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ According to question, $\mathrm{M}_{\mathrm{P}}=2 \mathrm{M}_{\mathrm{e}}$ $\mathrm{R}_{\mathrm{P}}=3 \mathrm{R}_{\mathrm{e}}$ Now, $\quad g_{P}=\frac{G \times 2 M_{e}}{\left(3 R_{e}\right)^{2}}$ $\mathrm{g}_{\mathrm{P}}=\frac{2 \mathrm{GM}_{\mathrm{e}}}{9 \mathrm{R}_{\mathrm{e}}^{2}}$ Divide equation (i) by equation (ii), we get- $\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{P}}}=\frac{9}{2}$ $\frac{10}{\mathrm{~g}_{\mathrm{P}}}=\frac{9}{2} \Rightarrow \mathrm{g}_{\mathrm{P}}=2.22 \mathrm{~m} / \mathrm{sec}^{2}$
AP EAMCET-07.09.2021
Gravitation
138391
At a distance $320 \mathrm{~km}$ above the surface of the earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly - (earth's radius $=6400$ km)
1 $2 \%$
2 $6 \%$
3 $10 \%$
4 $14 \%$
Explanation:
C Height above the surface $(\mathrm{h})=320 \mathrm{~km}$ As we know for smallest height, $\frac{g^{\prime}}{g}=\left(1-\frac{2 h}{R}\right)=\frac{9}{10}$ $g^{\prime}=\frac{9}{10} g$ $\therefore \% \text { decrease in } g=\frac{g-g^{\prime}}{g} \times 100$ $=\frac{g-\frac{9}{10} g}{g} \times 100$ $=10 \%$
