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PHXI08:GRAVITATION

359775 What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.
Given :
(i) The value of gravitational acceleration $g = 10 m s^{- 2}$
(ii) Radius of earth $R_{E} = 6400 k m .$
Take $π = 3.14$

1 83.73 minutes

2 85 minutes

3 90 minutes

4 156 minutes

Explanation:

Period of revolution of earth satellite is
$T = 2 π \sqrt{\frac{{(R_{E} + h)}^{3}}{g R_{E}^{2}}}$
As $h << R_{E}$
$T = 2 π \sqrt{\frac{R_{E}}{g}}$
Substituting the given values, we get
$T = 2 \times 3.14 \sqrt{\frac{6400 \times 10^{3}}{10}} s$
$= 5024 s = 83.73$ minutes

KCET - 2011

PHXI08:GRAVITATION

359776 If $g \propto \frac{1}{R^{3}} ($ instead of $\frac{1}{R^{2}}$ ), then the relation between time period of a satellite near earth's surface and radius $R$ will be

1

T \propto R^{2}

2

T \propto T

3

T^{2} \propto R

4

T^{2} \propto R^{2}

Explanation:

Gravitational force provides the required centripetal force
$\begin{aligned} m ω^{2} R = m g \\ m ω^{2} R = \frac{G M m}{R^{3}} \Rightarrow \frac{4 π^{2}}{T^{2}} = \frac{G M}{R^{4}} \Rightarrow T \propto R^{2} . \end{aligned}$

PHXI08:GRAVITATION

359777 The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is $v$ . For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is

1

\frac{3}{2} v

2

\sqrt{\frac{3}{2}} v

3

\sqrt{\frac{2}{3}} v

4

\frac{2}{3} v

Explanation:

$v = \sqrt{\frac{G M}{r}}$
or $v \propto \frac{1}{\sqrt{r}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{r_{1}}{r_{2}}}$
$∴ v_{2} = \sqrt{\frac{r_{1}}{r_{2}}} . v$
$= \sqrt{\frac{R}{R + \frac{R}{2}}} \cdot v = \sqrt{\frac{2}{3}} v$

PHXI08:GRAVITATION

359778 Assertion :
Escape velocity is greater than its orbital velocity of a satellite.
Reason :
Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v_{e} = \sqrt{2} v_{0} \Rightarrow v_{e} > v_{0}$
$v_{e} = 11.2 k m / s$
So correct option is (1).

PHXI08:GRAVITATION

359775 What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.
Given :
(i) The value of gravitational acceleration $g = 10 m s^{- 2}$
(ii) Radius of earth $R_{E} = 6400 k m .$
Take $π = 3.14$

1 83.73 minutes

2 85 minutes

3 90 minutes

4 156 minutes

Explanation:

Period of revolution of earth satellite is
$T = 2 π \sqrt{\frac{{(R_{E} + h)}^{3}}{g R_{E}^{2}}}$
As $h << R_{E}$
$T = 2 π \sqrt{\frac{R_{E}}{g}}$
Substituting the given values, we get
$T = 2 \times 3.14 \sqrt{\frac{6400 \times 10^{3}}{10}} s$
$= 5024 s = 83.73$ minutes

KCET - 2011

PHXI08:GRAVITATION

359776 If $g \propto \frac{1}{R^{3}} ($ instead of $\frac{1}{R^{2}}$ ), then the relation between time period of a satellite near earth's surface and radius $R$ will be

1

T \propto R^{2}

2

T \propto T

3

T^{2} \propto R

4

T^{2} \propto R^{2}

Explanation:

Gravitational force provides the required centripetal force
$\begin{aligned} m ω^{2} R = m g \\ m ω^{2} R = \frac{G M m}{R^{3}} \Rightarrow \frac{4 π^{2}}{T^{2}} = \frac{G M}{R^{4}} \Rightarrow T \propto R^{2} . \end{aligned}$

PHXI08:GRAVITATION

359777 The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is $v$ . For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is

1

\frac{3}{2} v

2

\sqrt{\frac{3}{2}} v

3

\sqrt{\frac{2}{3}} v

4

\frac{2}{3} v

Explanation:

$v = \sqrt{\frac{G M}{r}}$
or $v \propto \frac{1}{\sqrt{r}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{r_{1}}{r_{2}}}$
$∴ v_{2} = \sqrt{\frac{r_{1}}{r_{2}}} . v$
$= \sqrt{\frac{R}{R + \frac{R}{2}}} \cdot v = \sqrt{\frac{2}{3}} v$

PHXI08:GRAVITATION

359778 Assertion :
Escape velocity is greater than its orbital velocity of a satellite.
Reason :
Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v_{e} = \sqrt{2} v_{0} \Rightarrow v_{e} > v_{0}$
$v_{e} = 11.2 k m / s$
So correct option is (1).

PHXI08:GRAVITATION

359775 What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.
Given :
(i) The value of gravitational acceleration $g = 10 m s^{- 2}$
(ii) Radius of earth $R_{E} = 6400 k m .$
Take $π = 3.14$

1 83.73 minutes

2 85 minutes

3 90 minutes

4 156 minutes

Explanation:

Period of revolution of earth satellite is
$T = 2 π \sqrt{\frac{{(R_{E} + h)}^{3}}{g R_{E}^{2}}}$
As $h << R_{E}$
$T = 2 π \sqrt{\frac{R_{E}}{g}}$
Substituting the given values, we get
$T = 2 \times 3.14 \sqrt{\frac{6400 \times 10^{3}}{10}} s$
$= 5024 s = 83.73$ minutes

KCET - 2011

PHXI08:GRAVITATION

359776 If $g \propto \frac{1}{R^{3}} ($ instead of $\frac{1}{R^{2}}$ ), then the relation between time period of a satellite near earth's surface and radius $R$ will be

1

T \propto R^{2}

2

T \propto T

3

T^{2} \propto R

4

T^{2} \propto R^{2}

Explanation:

Gravitational force provides the required centripetal force
$\begin{aligned} m ω^{2} R = m g \\ m ω^{2} R = \frac{G M m}{R^{3}} \Rightarrow \frac{4 π^{2}}{T^{2}} = \frac{G M}{R^{4}} \Rightarrow T \propto R^{2} . \end{aligned}$

PHXI08:GRAVITATION

359777 The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is $v$ . For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is

1

\frac{3}{2} v

2

\sqrt{\frac{3}{2}} v

3

\sqrt{\frac{2}{3}} v

4

\frac{2}{3} v

Explanation:

$v = \sqrt{\frac{G M}{r}}$
or $v \propto \frac{1}{\sqrt{r}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{r_{1}}{r_{2}}}$
$∴ v_{2} = \sqrt{\frac{r_{1}}{r_{2}}} . v$
$= \sqrt{\frac{R}{R + \frac{R}{2}}} \cdot v = \sqrt{\frac{2}{3}} v$

PHXI08:GRAVITATION

359778 Assertion :
Escape velocity is greater than its orbital velocity of a satellite.
Reason :
Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v_{e} = \sqrt{2} v_{0} \Rightarrow v_{e} > v_{0}$
$v_{e} = 11.2 k m / s$
So correct option is (1).

PHXI08:GRAVITATION

359775 What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.
Given :
(i) The value of gravitational acceleration $g = 10 m s^{- 2}$
(ii) Radius of earth $R_{E} = 6400 k m .$
Take $π = 3.14$

1 83.73 minutes

2 85 minutes

3 90 minutes

4 156 minutes

Explanation:

Period of revolution of earth satellite is
$T = 2 π \sqrt{\frac{{(R_{E} + h)}^{3}}{g R_{E}^{2}}}$
As $h << R_{E}$
$T = 2 π \sqrt{\frac{R_{E}}{g}}$
Substituting the given values, we get
$T = 2 \times 3.14 \sqrt{\frac{6400 \times 10^{3}}{10}} s$
$= 5024 s = 83.73$ minutes

KCET - 2011

PHXI08:GRAVITATION

359776 If $g \propto \frac{1}{R^{3}} ($ instead of $\frac{1}{R^{2}}$ ), then the relation between time period of a satellite near earth's surface and radius $R$ will be

1

T \propto R^{2}

2

T \propto T

3

T^{2} \propto R

4

T^{2} \propto R^{2}

Explanation:

Gravitational force provides the required centripetal force
$\begin{aligned} m ω^{2} R = m g \\ m ω^{2} R = \frac{G M m}{R^{3}} \Rightarrow \frac{4 π^{2}}{T^{2}} = \frac{G M}{R^{4}} \Rightarrow T \propto R^{2} . \end{aligned}$

PHXI08:GRAVITATION

359777 The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is $v$ . For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is

1

\frac{3}{2} v

2

\sqrt{\frac{3}{2}} v

3

\sqrt{\frac{2}{3}} v

4

\frac{2}{3} v

Explanation:

$v = \sqrt{\frac{G M}{r}}$
or $v \propto \frac{1}{\sqrt{r}}$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{r_{1}}{r_{2}}}$
$∴ v_{2} = \sqrt{\frac{r_{1}}{r_{2}}} . v$
$= \sqrt{\frac{R}{R + \frac{R}{2}}} \cdot v = \sqrt{\frac{2}{3}} v$

PHXI08:GRAVITATION

359778 Assertion :
Escape velocity is greater than its orbital velocity of a satellite.
Reason :
Orbit of a satellite is within the gravitational field of earth whereas escaping is beyond the gravitational field of earth.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$v_{e} = \sqrt{2} v_{0} \Rightarrow v_{e} > v_{0}$
$v_{e} = 11.2 k m / s$
So correct option is (1).

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