Fg-6-67- 10-11Nm2-kg2- 6- 1024kg- 1kg6-4- 106m2-

F=6.67×10^ 11Nm^2/kg^2×6×10^24kg×1 kg(6.4× 10^6 m)^2 F=6.67 × 10^ 11Nm^2/kg^2×6×10^24kg×1kg(6.4×10^6m)^2 F=6.67×6×10^ 11+24(6.4×10^6)^2 F=40 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VI

Mathematics

Estimating Sum or Difference

Fg=6.67× 10-1...

Question

$F g = \frac{6.67 \times 10^{- 11} N m^{2} / k g^{2} \times 6 \times 10^{24} k g \times 1 k g}{(6.4 \times 10^{6} m)^{2}}$ .

Open in App

Solution

$F = \frac{6.67 \times 10^{- 11} N m^{2} / k g^{2} \times 6 \times 10^{24} k g \times 1 k g}{(6.4 \times 10^{6} m)^{2}}$
$F = \frac{6.67 \times 10^{- 11} \frac{N m^{2}}{k g^{2}} \times 6 \times 10^{24} k g \times 1 k g}{(6.4 \times 10^{6} m)^{2}}$
$F = \frac{6.67 \times 6 \times 10^{- 11 + 24}}{(6.4 \times 10^{6})^{2}}$
$F = \frac{40.02 \times 10^{13}}{40.96 \times 10^{12}}$
$F = 0.977050781 \times 10$
$F = 9.77050781$ Ans.

Suggest Corrections

Similar questions

Q. Calculate the height of the communication satellite.

[G i v e n G = 6.67 \times 10^{- 11} N m^{2} / k g^{2}, M = 6 \times 10^{24} k g, R = 6400 k m]

Q. Calculate the height of the communication satellite. [Given

G = 6.67 \times 10^{- 11} N m^{2} / k g^{2}, M = 6 \times 10^{24} k g, R = 6400 k m

(Ans:

35.9 \times 10^{6} m

)

Q. Find the binding energy of a body of mass

50

kg at rest on the surface of the earth.
given

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

R = 6400 k m, M = 6 \times 10^{24} k g

Q. Find the total energy and binding energy of an artificial satellite of mass

1000 k g

orbiting at height of

1600 k m

above the earth's surface. Given

G = 6.67 \times 10^{- 11} N m^{2} / k g^{2}, R = 6400 k m, M = 6 \times 10^{24} k g

What is the force of gravity on a body of mass 150 kg lying on the surface of the earth?
(Mass of earth = $6 \times 10^{24}$ kg; Radius of earth = $6.4 \times 10^{6} m; G = 6.7 \times 10^{- 11} N m^{2} / k g^{2}$ )

Join BYJU'S Learning Program

Fg=6.67×10−11Nm2/kg2×6×1024kg×1kg(6.4×106m)2.

F=6.67×10−11Nm2/kg2×6×1024kg×1kg(6.4×106m)2F=6.67×10−11Nm2kg2×6×1024kg×1kg(6.4×106m)2F=6.67×6×10−11+24(6.4×106)2F=40.02×101340.96×1012F=0.977050781×10F=9.77050781 Ans.

$F g = \frac{6.67 \times 10^{- 11} N m^{2} / k g^{2} \times 6 \times 10^{24} k g \times 1 k g}{(6.4 \times 10^{6} m)^{2}}$ .