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Standard XII

Physics

Variation of G Due to the Rotation of Earth

Mass of earth...

Question

Mass of earth when it's radius is $6400 k m$ and gravitational field strength is $9.81 N k g^{- 1}$ is

6.0 \times 10^{24} k g

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5 \times 10^{23} k g

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40 \times 10^{9} k g

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9 \times 10^{2} 4 k g

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Solution

The correct option is A $6.0 \times 10^{24} k g$
$G$ is gravitational constant = $6.67 \times 10^{- 11} u n i t s$
$R$ is radius of earth
$g$ is gravitational field strength

$g =$ $\frac{G M}{R^{2}}$
$⟹ M = \frac{g R^{2}}{G}$
$⟹ M = 6 \times 10^{24} k g$

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Similar questions

Q. The distance between the earth and the moon is

3.85 \times 10^{8}

metre.At what point in between the two will the gravitational field intensity be zero? Mass of the earth

= 6.0 \times 10^{24} k g

, mass of the moon

= 7.26 \times 10^{22} k g

The distance of a point from the earth's centre where the resultant gravitational field due to the earth and the moon is zero is...( The mass of the earth is 6.0 $\times$ $10^{24}$ kg and that of the moon is 7.4 $\times$ $10^{22}$ kg. The distance between the earth and the moon is 40 $\times$ $10^{5}$ km.

Q. The mass of the earth is

6 \times 10^{24} k g

. The distance between the earth and the Sun is

1.5 \times 10^{11} m

. If the gravitational force between the two is

3.5 \times 10^{22} N

, what is the mass of the Sun? Use

G = 6.67 \times 10^{- 11} N m^{2} k g^{- 2}

Question 3
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is $6 \times 10^{24} k g$ and radius of the earth is $6.4 \times 10^{6} m$ )

Q. A satellite of mass

250 k g

is orbiting the Earth a height of

500 k m

above the surface of Earth. How much energy must be expended to rocket the satellite out of the gravitational influence of the Earth ? Given mass of the Earth

= 6.0 \times 10^{24} k g

radius of the Earth = 6400 km ; and G = 6.67 x

10^{- 11} N m^{2} k g^{- 2}

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Mass of earth when it's radius is 6400km and gravitational field strength is 9.81Nkg−1 is

The correct option is A 6.0×1024kgG is gravitational constant = 6.67×10−11units R is radius of earthg is gravitational field strengthg= GMR2⟹M=gR2G⟹M=6×1024kg

Mass of earth when it's radius is $6400 k m$ and gravitational field strength is $9.81 N k g^{- 1}$ is

The correct option is A $6.0 \times 10^{24} k g$
$G$ is gravitational constant = $6.67 \times 10^{- 11} u n i t s$
$R$ is radius of earth
$g$ is gravitational field strength

$g =$ $\frac{G M}{R^{2}}$
$⟹ M = \frac{g R^{2}}{G}$
$⟹ M = 6 \times 10^{24} k g$