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PHXI08:GRAVITATION

359675 If the radius of earth is 6000 $k m$ , what will be the weight of 120 $k g$ body if taken to a height of 2000 $k m$ above sea level (i.e., surface of the Earth).

1

67.5 k g w t

2

38.2 k g w t

3

56.2 k g w t

4

78.1 k g w t

Explanation:

With usual notation, $R = 6000 k m, h = 2000 k m$
If $g$ and $g_{h}$ , denote the values of acceleration due to gravity on the surface and at a height $h$ ,then
$\frac{g_{h}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{{(6000)}^{2}}{{(6000 + 2000)}^{2}}$
$= \frac{36 \times 10^{6}}{64 \times 10^{6}} o r \frac{g_{h}}{g} = \frac{9}{16}$
Thus, the weight at a height of 2000 km
$= (\frac{9}{16}) (120 k g) = 67.5 k g w t$

PHXI08:GRAVITATION

359676 At what height above the surface of earth, the value of acceleration due to gravity would be half of its value on the surface of earth? (take, radius of the earth is $6400 k m$ )

1

2561 k m

2

2650 k m

3

3200 k m

4

9800 k m

Explanation:

The ratio,
$\frac{g^{'}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{1}{2}$
$\Rightarrow R + h = \sqrt{2} R$
$\Rightarrow h = (\sqrt{2} - 1) R$
$\Rightarrow h = (0.414) \times 6400 (g i v e n)$
$∴ h = 2650 k m$

AIIMS - 2009

PHXI08:GRAVITATION

359677 What is the value of acceleration due to gravity at a height equal to half the radius of the Earth, from its surface ?

1

4.4 {m s}^{- 2}

2

6.5 {m s}^{- 2}

3 zero

4

9.8 {m s}^{- 2}

Explanation:

Acceleration due to gravity at a height is given by
$g_{h} = \frac{g R^{2}}{(R + h)^{2}} = \frac{4 g}{9} = \frac{4}{9} \times 9.8 = 4.4 m / s^{2}$

KCET - 2024

PHXI08:GRAVITATION

359678 The weight of a body on the surface of the earth is $63 N$ . What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

1

35 N

2

28 N

3

18 N

4

40 N

Explanation:

Given, $h = \frac{R_{e}}{2}$
Acceleration due to gravity at altitude $h$ is given by
$\begin{aligned} g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}} = \frac{g}{{(1 + \frac{R_{e} / 2}{R_{e}})}^{2}} \\ = \frac{g}{{(1 + \frac{1}{2})}^{2}} = \frac{g}{{(\frac{3}{2})}^{2}} = \frac{4 g}{9} \end{aligned}$
$\Rightarrow w^{1} = \frac{4}{9} \times 63$
$= 28 N$

AIIMS - 2018

PHXI08:GRAVITATION

359675 If the radius of earth is 6000 $k m$ , what will be the weight of 120 $k g$ body if taken to a height of 2000 $k m$ above sea level (i.e., surface of the Earth).

1

67.5 k g w t

2

38.2 k g w t

3

56.2 k g w t

4

78.1 k g w t

Explanation:

With usual notation, $R = 6000 k m, h = 2000 k m$
If $g$ and $g_{h}$ , denote the values of acceleration due to gravity on the surface and at a height $h$ ,then
$\frac{g_{h}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{{(6000)}^{2}}{{(6000 + 2000)}^{2}}$
$= \frac{36 \times 10^{6}}{64 \times 10^{6}} o r \frac{g_{h}}{g} = \frac{9}{16}$
Thus, the weight at a height of 2000 km
$= (\frac{9}{16}) (120 k g) = 67.5 k g w t$

PHXI08:GRAVITATION

359676 At what height above the surface of earth, the value of acceleration due to gravity would be half of its value on the surface of earth? (take, radius of the earth is $6400 k m$ )

1

2561 k m

2

2650 k m

3

3200 k m

4

9800 k m

Explanation:

The ratio,
$\frac{g^{'}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{1}{2}$
$\Rightarrow R + h = \sqrt{2} R$
$\Rightarrow h = (\sqrt{2} - 1) R$
$\Rightarrow h = (0.414) \times 6400 (g i v e n)$
$∴ h = 2650 k m$

AIIMS - 2009

PHXI08:GRAVITATION

359677 What is the value of acceleration due to gravity at a height equal to half the radius of the Earth, from its surface ?

1

4.4 {m s}^{- 2}

2

6.5 {m s}^{- 2}

3 zero

4

9.8 {m s}^{- 2}

Explanation:

Acceleration due to gravity at a height is given by
$g_{h} = \frac{g R^{2}}{(R + h)^{2}} = \frac{4 g}{9} = \frac{4}{9} \times 9.8 = 4.4 m / s^{2}$

KCET - 2024

PHXI08:GRAVITATION

359678 The weight of a body on the surface of the earth is $63 N$ . What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

1

35 N

2

28 N

3

18 N

4

40 N

Explanation:

Given, $h = \frac{R_{e}}{2}$
Acceleration due to gravity at altitude $h$ is given by
$\begin{aligned} g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}} = \frac{g}{{(1 + \frac{R_{e} / 2}{R_{e}})}^{2}} \\ = \frac{g}{{(1 + \frac{1}{2})}^{2}} = \frac{g}{{(\frac{3}{2})}^{2}} = \frac{4 g}{9} \end{aligned}$
$\Rightarrow w^{1} = \frac{4}{9} \times 63$
$= 28 N$

AIIMS - 2018

PHXI08:GRAVITATION

359675 If the radius of earth is 6000 $k m$ , what will be the weight of 120 $k g$ body if taken to a height of 2000 $k m$ above sea level (i.e., surface of the Earth).

1

67.5 k g w t

2

38.2 k g w t

3

56.2 k g w t

4

78.1 k g w t

Explanation:

With usual notation, $R = 6000 k m, h = 2000 k m$
If $g$ and $g_{h}$ , denote the values of acceleration due to gravity on the surface and at a height $h$ ,then
$\frac{g_{h}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{{(6000)}^{2}}{{(6000 + 2000)}^{2}}$
$= \frac{36 \times 10^{6}}{64 \times 10^{6}} o r \frac{g_{h}}{g} = \frac{9}{16}$
Thus, the weight at a height of 2000 km
$= (\frac{9}{16}) (120 k g) = 67.5 k g w t$

PHXI08:GRAVITATION

359676 At what height above the surface of earth, the value of acceleration due to gravity would be half of its value on the surface of earth? (take, radius of the earth is $6400 k m$ )

1

2561 k m

2

2650 k m

3

3200 k m

4

9800 k m

Explanation:

The ratio,
$\frac{g^{'}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{1}{2}$
$\Rightarrow R + h = \sqrt{2} R$
$\Rightarrow h = (\sqrt{2} - 1) R$
$\Rightarrow h = (0.414) \times 6400 (g i v e n)$
$∴ h = 2650 k m$

AIIMS - 2009

PHXI08:GRAVITATION

359677 What is the value of acceleration due to gravity at a height equal to half the radius of the Earth, from its surface ?

1

4.4 {m s}^{- 2}

2

6.5 {m s}^{- 2}

3 zero

4

9.8 {m s}^{- 2}

Explanation:

Acceleration due to gravity at a height is given by
$g_{h} = \frac{g R^{2}}{(R + h)^{2}} = \frac{4 g}{9} = \frac{4}{9} \times 9.8 = 4.4 m / s^{2}$

KCET - 2024

PHXI08:GRAVITATION

359678 The weight of a body on the surface of the earth is $63 N$ . What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

1

35 N

2

28 N

3

18 N

4

40 N

Explanation:

Given, $h = \frac{R_{e}}{2}$
Acceleration due to gravity at altitude $h$ is given by
$\begin{aligned} g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}} = \frac{g}{{(1 + \frac{R_{e} / 2}{R_{e}})}^{2}} \\ = \frac{g}{{(1 + \frac{1}{2})}^{2}} = \frac{g}{{(\frac{3}{2})}^{2}} = \frac{4 g}{9} \end{aligned}$
$\Rightarrow w^{1} = \frac{4}{9} \times 63$
$= 28 N$

AIIMS - 2018

PHXI08:GRAVITATION

359675 If the radius of earth is 6000 $k m$ , what will be the weight of 120 $k g$ body if taken to a height of 2000 $k m$ above sea level (i.e., surface of the Earth).

1

67.5 k g w t

2

38.2 k g w t

3

56.2 k g w t

4

78.1 k g w t

Explanation:

With usual notation, $R = 6000 k m, h = 2000 k m$
If $g$ and $g_{h}$ , denote the values of acceleration due to gravity on the surface and at a height $h$ ,then
$\frac{g_{h}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{{(6000)}^{2}}{{(6000 + 2000)}^{2}}$
$= \frac{36 \times 10^{6}}{64 \times 10^{6}} o r \frac{g_{h}}{g} = \frac{9}{16}$
Thus, the weight at a height of 2000 km
$= (\frac{9}{16}) (120 k g) = 67.5 k g w t$

PHXI08:GRAVITATION

359676 At what height above the surface of earth, the value of acceleration due to gravity would be half of its value on the surface of earth? (take, radius of the earth is $6400 k m$ )

1

2561 k m

2

2650 k m

3

3200 k m

4

9800 k m

Explanation:

The ratio,
$\frac{g^{'}}{g} = \frac{R^{2}}{{(R + h)}^{2}} = \frac{1}{2}$
$\Rightarrow R + h = \sqrt{2} R$
$\Rightarrow h = (\sqrt{2} - 1) R$
$\Rightarrow h = (0.414) \times 6400 (g i v e n)$
$∴ h = 2650 k m$

AIIMS - 2009

PHXI08:GRAVITATION

359677 What is the value of acceleration due to gravity at a height equal to half the radius of the Earth, from its surface ?

1

4.4 {m s}^{- 2}

2

6.5 {m s}^{- 2}

3 zero

4

9.8 {m s}^{- 2}

Explanation:

Acceleration due to gravity at a height is given by
$g_{h} = \frac{g R^{2}}{(R + h)^{2}} = \frac{4 g}{9} = \frac{4}{9} \times 9.8 = 4.4 m / s^{2}$

KCET - 2024

PHXI08:GRAVITATION

359678 The weight of a body on the surface of the earth is $63 N$ . What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

1

35 N

2

28 N

3

18 N

4

40 N

Explanation:

Given, $h = \frac{R_{e}}{2}$
Acceleration due to gravity at altitude $h$ is given by
$\begin{aligned} g^{'} = \frac{g}{{(1 + \frac{h}{R_{e}})}^{2}} = \frac{g}{{(1 + \frac{R_{e} / 2}{R_{e}})}^{2}} \\ = \frac{g}{{(1 + \frac{1}{2})}^{2}} = \frac{g}{{(\frac{3}{2})}^{2}} = \frac{4 g}{9} \end{aligned}$
$\Rightarrow w^{1} = \frac{4}{9} \times 63$
$= 28 N$

AIIMS - 2018

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