138761
A rocket is fired from inside a deep mine, so as to escape the earth's gravitational field. The minimum velocity to be imparted to the rocket is
1 exactly the same as the escape velocity of fire from the earth's surface.
2 a little more than the escape velocity of fire from the earth's surface.
3 a little less than the escape velocity of fire from the earth's surface.
4 infinity
Explanation:
B Since, escape velocity, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore$ Velocity imparted to the rocket, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}-\mathrm{h}}}$ $\therefore \mathrm{v}_{\mathrm{e}}{ }^{\prime}>\mathrm{v}_{\mathrm{e}}$ i.e. A little more than the escape velocity of fire from the earth's surface
J and K-CET-2012
Gravitation
138762
How long will a satellite, placed in a circular orbit of radius that is $\left(\frac{1}{4}\right)^{\text {th }}$ the radius of a geostationary satellite, take to complete one revolution around the earth?
1 12 hours
2 6 hours
3 3 hours
4 4 days
Explanation:
C According to Kepler's third law or law of period, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}_{1}}{\mathrm{R}_{1} / 4}\right]^{3} \quad\left\{\because \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{4}\right\}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=64$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]=8$ $\mathrm{T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}$ $\therefore \quad \mathrm{T}_{2}=3$ hours$
J and K-CET-2012
Gravitation
138763
Assuming density $d$ of a planet to be uniform, we can say that the time period of its artificial satellite is proportional to
1 $d$
2 $\sqrt{\mathrm{d}}$
3 $\frac{1}{\sqrt{\mathrm{d}}}$
4 $\frac{1}{\mathrm{~d}}$
Explanation:
C According to question, Density of planet $(\mathrm{d})=\frac{\operatorname{Mass}(\mathrm{m})}{\operatorname{Volume}(\mathrm{V})}$ $\mathrm{d}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^{3}}$ $\mathrm{d} \propto \frac{1}{\mathrm{r}^{3}}$ $\mathrm{r}^{3} \propto \frac{1}{\mathrm{~d}}$ Then, from Kepler's third law, $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ $\mathrm{~T}^{2} \propto \frac{1}{\mathrm{~d}}$ $\therefore \quad \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{d}}}$
J and K-CET-2015
Gravitation
138764
A satellite of mass ' $m$ ', revolving round the earth of radius ' $r$ ' has kinetic energy $(E)$. Its angular momentum is
1 $\left(\mathrm{mEr}^{2}\right)$
2 $\left(2 \mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
3 $\left(\mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
4 $\left(2 \mathrm{mEr}^{2}\right)$
Explanation:
B Given that, Mass of satellite $=\mathrm{m}$ Radius of earth $=r$ Kinetic energy $=\mathrm{E}$ Angular momentum $=$ ? Kinetic Energy $(E)=\frac{1}{2} \mathrm{mv}^2$ $\mathrm{v}^2=\frac{2 \mathrm{E}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}}$ We know, Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m} \times \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \times \mathrm{r}$ $\mathrm{L}=\sqrt{2 \mathrm{mEr}^2}$ $\mathrm{~L}=\left(2 \mathrm{mEr}^2\right)^{1 / 2}$
138761
A rocket is fired from inside a deep mine, so as to escape the earth's gravitational field. The minimum velocity to be imparted to the rocket is
1 exactly the same as the escape velocity of fire from the earth's surface.
2 a little more than the escape velocity of fire from the earth's surface.
3 a little less than the escape velocity of fire from the earth's surface.
4 infinity
Explanation:
B Since, escape velocity, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore$ Velocity imparted to the rocket, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}-\mathrm{h}}}$ $\therefore \mathrm{v}_{\mathrm{e}}{ }^{\prime}>\mathrm{v}_{\mathrm{e}}$ i.e. A little more than the escape velocity of fire from the earth's surface
J and K-CET-2012
Gravitation
138762
How long will a satellite, placed in a circular orbit of radius that is $\left(\frac{1}{4}\right)^{\text {th }}$ the radius of a geostationary satellite, take to complete one revolution around the earth?
1 12 hours
2 6 hours
3 3 hours
4 4 days
Explanation:
C According to Kepler's third law or law of period, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}_{1}}{\mathrm{R}_{1} / 4}\right]^{3} \quad\left\{\because \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{4}\right\}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=64$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]=8$ $\mathrm{T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}$ $\therefore \quad \mathrm{T}_{2}=3$ hours$
J and K-CET-2012
Gravitation
138763
Assuming density $d$ of a planet to be uniform, we can say that the time period of its artificial satellite is proportional to
1 $d$
2 $\sqrt{\mathrm{d}}$
3 $\frac{1}{\sqrt{\mathrm{d}}}$
4 $\frac{1}{\mathrm{~d}}$
Explanation:
C According to question, Density of planet $(\mathrm{d})=\frac{\operatorname{Mass}(\mathrm{m})}{\operatorname{Volume}(\mathrm{V})}$ $\mathrm{d}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^{3}}$ $\mathrm{d} \propto \frac{1}{\mathrm{r}^{3}}$ $\mathrm{r}^{3} \propto \frac{1}{\mathrm{~d}}$ Then, from Kepler's third law, $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ $\mathrm{~T}^{2} \propto \frac{1}{\mathrm{~d}}$ $\therefore \quad \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{d}}}$
J and K-CET-2015
Gravitation
138764
A satellite of mass ' $m$ ', revolving round the earth of radius ' $r$ ' has kinetic energy $(E)$. Its angular momentum is
1 $\left(\mathrm{mEr}^{2}\right)$
2 $\left(2 \mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
3 $\left(\mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
4 $\left(2 \mathrm{mEr}^{2}\right)$
Explanation:
B Given that, Mass of satellite $=\mathrm{m}$ Radius of earth $=r$ Kinetic energy $=\mathrm{E}$ Angular momentum $=$ ? Kinetic Energy $(E)=\frac{1}{2} \mathrm{mv}^2$ $\mathrm{v}^2=\frac{2 \mathrm{E}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}}$ We know, Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m} \times \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \times \mathrm{r}$ $\mathrm{L}=\sqrt{2 \mathrm{mEr}^2}$ $\mathrm{~L}=\left(2 \mathrm{mEr}^2\right)^{1 / 2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138761
A rocket is fired from inside a deep mine, so as to escape the earth's gravitational field. The minimum velocity to be imparted to the rocket is
1 exactly the same as the escape velocity of fire from the earth's surface.
2 a little more than the escape velocity of fire from the earth's surface.
3 a little less than the escape velocity of fire from the earth's surface.
4 infinity
Explanation:
B Since, escape velocity, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore$ Velocity imparted to the rocket, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}-\mathrm{h}}}$ $\therefore \mathrm{v}_{\mathrm{e}}{ }^{\prime}>\mathrm{v}_{\mathrm{e}}$ i.e. A little more than the escape velocity of fire from the earth's surface
J and K-CET-2012
Gravitation
138762
How long will a satellite, placed in a circular orbit of radius that is $\left(\frac{1}{4}\right)^{\text {th }}$ the radius of a geostationary satellite, take to complete one revolution around the earth?
1 12 hours
2 6 hours
3 3 hours
4 4 days
Explanation:
C According to Kepler's third law or law of period, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}_{1}}{\mathrm{R}_{1} / 4}\right]^{3} \quad\left\{\because \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{4}\right\}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=64$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]=8$ $\mathrm{T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}$ $\therefore \quad \mathrm{T}_{2}=3$ hours$
J and K-CET-2012
Gravitation
138763
Assuming density $d$ of a planet to be uniform, we can say that the time period of its artificial satellite is proportional to
1 $d$
2 $\sqrt{\mathrm{d}}$
3 $\frac{1}{\sqrt{\mathrm{d}}}$
4 $\frac{1}{\mathrm{~d}}$
Explanation:
C According to question, Density of planet $(\mathrm{d})=\frac{\operatorname{Mass}(\mathrm{m})}{\operatorname{Volume}(\mathrm{V})}$ $\mathrm{d}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^{3}}$ $\mathrm{d} \propto \frac{1}{\mathrm{r}^{3}}$ $\mathrm{r}^{3} \propto \frac{1}{\mathrm{~d}}$ Then, from Kepler's third law, $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ $\mathrm{~T}^{2} \propto \frac{1}{\mathrm{~d}}$ $\therefore \quad \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{d}}}$
J and K-CET-2015
Gravitation
138764
A satellite of mass ' $m$ ', revolving round the earth of radius ' $r$ ' has kinetic energy $(E)$. Its angular momentum is
1 $\left(\mathrm{mEr}^{2}\right)$
2 $\left(2 \mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
3 $\left(\mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
4 $\left(2 \mathrm{mEr}^{2}\right)$
Explanation:
B Given that, Mass of satellite $=\mathrm{m}$ Radius of earth $=r$ Kinetic energy $=\mathrm{E}$ Angular momentum $=$ ? Kinetic Energy $(E)=\frac{1}{2} \mathrm{mv}^2$ $\mathrm{v}^2=\frac{2 \mathrm{E}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}}$ We know, Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m} \times \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \times \mathrm{r}$ $\mathrm{L}=\sqrt{2 \mathrm{mEr}^2}$ $\mathrm{~L}=\left(2 \mathrm{mEr}^2\right)^{1 / 2}$
138761
A rocket is fired from inside a deep mine, so as to escape the earth's gravitational field. The minimum velocity to be imparted to the rocket is
1 exactly the same as the escape velocity of fire from the earth's surface.
2 a little more than the escape velocity of fire from the earth's surface.
3 a little less than the escape velocity of fire from the earth's surface.
4 infinity
Explanation:
B Since, escape velocity, $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$ $\therefore$ Velocity imparted to the rocket, $\mathrm{v}_{\mathrm{e}}{ }^{\prime}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}-\mathrm{h}}}$ $\therefore \mathrm{v}_{\mathrm{e}}{ }^{\prime}>\mathrm{v}_{\mathrm{e}}$ i.e. A little more than the escape velocity of fire from the earth's surface
J and K-CET-2012
Gravitation
138762
How long will a satellite, placed in a circular orbit of radius that is $\left(\frac{1}{4}\right)^{\text {th }}$ the radius of a geostationary satellite, take to complete one revolution around the earth?
1 12 hours
2 6 hours
3 3 hours
4 4 days
Explanation:
C According to Kepler's third law or law of period, $\mathrm{T}^{2} \propto \mathrm{R}^{3}$ $\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\mathrm{R}_{1}^{3}}{\mathrm{R}_{2}^{3}}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=\left[\frac{\mathrm{R}_{1}}{\mathrm{R}_{1} / 4}\right]^{3} \quad\left\{\because \mathrm{R}_{2}=\frac{\mathrm{R}_{1}}{4}\right\}$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{2}=64$ $\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]=8$ $\mathrm{T}_{2}=\frac{\mathrm{T}_{1}}{8}=\frac{24}{8}$ $\therefore \quad \mathrm{T}_{2}=3$ hours$
J and K-CET-2012
Gravitation
138763
Assuming density $d$ of a planet to be uniform, we can say that the time period of its artificial satellite is proportional to
1 $d$
2 $\sqrt{\mathrm{d}}$
3 $\frac{1}{\sqrt{\mathrm{d}}}$
4 $\frac{1}{\mathrm{~d}}$
Explanation:
C According to question, Density of planet $(\mathrm{d})=\frac{\operatorname{Mass}(\mathrm{m})}{\operatorname{Volume}(\mathrm{V})}$ $\mathrm{d}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{r}^{3}}$ $\mathrm{d} \propto \frac{1}{\mathrm{r}^{3}}$ $\mathrm{r}^{3} \propto \frac{1}{\mathrm{~d}}$ Then, from Kepler's third law, $\mathrm{T}^{2} \propto \mathrm{r}^{3}$ $\mathrm{~T}^{2} \propto \frac{1}{\mathrm{~d}}$ $\therefore \quad \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{d}}}$
J and K-CET-2015
Gravitation
138764
A satellite of mass ' $m$ ', revolving round the earth of radius ' $r$ ' has kinetic energy $(E)$. Its angular momentum is
1 $\left(\mathrm{mEr}^{2}\right)$
2 $\left(2 \mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
3 $\left(\mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
4 $\left(2 \mathrm{mEr}^{2}\right)$
Explanation:
B Given that, Mass of satellite $=\mathrm{m}$ Radius of earth $=r$ Kinetic energy $=\mathrm{E}$ Angular momentum $=$ ? Kinetic Energy $(E)=\frac{1}{2} \mathrm{mv}^2$ $\mathrm{v}^2=\frac{2 \mathrm{E}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}}$ We know, Angular momentum $(\mathrm{L})=\mathrm{mvr}$ $\mathrm{L}=\mathrm{m} \times \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \times \mathrm{r}$ $\mathrm{L}=\sqrt{2 \mathrm{mEr}^2}$ $\mathrm{~L}=\left(2 \mathrm{mEr}^2\right)^{1 / 2}$