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MAXImum [283]
3 years ago
6

Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?

Physics
2 answers:
Sever21 [200]3 years ago
5 0
<span>suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2                             ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2</span>
snow_tiger [21]3 years ago
4 0
Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2                             ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2




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torisob [31]

Answer:

The electric potential in volts is 1.618 x 10⁻¹⁷ V

Explanation:

The electric potential, in volts, at point P, can be calculated as follow;

Electric potential is the work done in moving a unit positive charge from infinity to a particular point in the electric field.

Thus, the work done in this process in moving the charge to point p is 101eV.

Convert this Volts = 101 × 1.602 x 10⁻¹⁹ V

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Therefore, the electric potential in volts is 1.618 x 10⁻¹⁷ V

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