Assuming the earth of to be a uniform sphere of radius `6400 km` and density `5.5 g//c.c.`, find the value of `g` on its surface`G=6.66xx10^(-11)Nm^(2)kg^(-2)`
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To find the value of acceleration due to gravity (g) on the surface of the Earth, we can follow these steps:
### Step 1: Understand the formula for acceleration due to gravity
The formula for acceleration due to gravity (g) at the surface of a sphere is given by:
\[ g = \frac{G \cdot M}{R^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the sphere (Earth in this case),
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Determine the escape velocity of a body from the moon. Take the moon to be a uniform sphere of radius 1.74 xx 10^(6) m , and the mass 7.36 xx 10^(22) kg ? ( G = 6.67 xx 10^(-11) Nm^(2) kg^(-2))
Assuming the earth b to be a homogeneous sphere determine the density of the earth from the following data. g = 9.8 m//s^(2), G = 6.67 xx 10^(-11) Nm^(2) kg^(2) , radius of the earth = 6372km.
Taking the earth to be a uniform sphere of radius 6400 km and the value of g at the surface to be 10ms^(-2) , calculate the energy needed to raise a satellite of mass 2000 kg to a height of 800 km a above the earth's surface and to set it into circular orbit at that altitude.
An artificial satellite is revolving in a circular orbit at a height of 1200 km above the surface of the earth. If the radius of the earth is 6400 km, mass is 6 xx 10^(24) kg find the orbital velocity (G = 6.67 xx 10^(-11)Nm^(2)//kg^(2))
Assuming the earth to be a homogeneous sphere of radius R , its density in terms of G (constant of gravitation) and g (acceleration due to gravity on the surface of the earth)
Mass of moon is 7.349 xx 10^(22) kg and its radius is 1.738 xx 10^(6) m . Calculate its mean density and acceleration due to gravity on its surface. Given G = 6668 xx 10^(-11) Nm^(2) kg ^(-2) .
If the acceleration due to gravity on earth is 9.81 m//s^(2) and the radius of the earth is 6370 km find the mass of the earth ? (G = 6.67 xx 10^(-11) Nm^(2)//kg^(2))
A satellite of mass 50kg orbits the earth at a height of 100km. Calcualte (i) K.E. (ii) P.E. and (iii) total energy . Given G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) . Radius of the arth is 6400 and mass of the earth is 6 xx 10^(24) kg.
Assuming earth to be a sphere of radius 6400km, calculate the height above the earth's surface at which the value of acceleration due to gravity reduces to half its value on the erath's surface.
The mass of the earth is 6 × 10^(24) kg and that of the moon is 7.4 xx 10^(22) kg. If the distance between the earth and the moon is 3.84xx10^(5) km, calculate the force exerted by the earth on the moon. (Take G = 6.7 xx 10^(–11) N m^(2) kg^(-2) )
