Question 127: Gravitation

Question

If the gravitational field in the space is given as $$\left(-\frac{K}{r^{2}}\right)$$. Taking the reference point to be at $$\mathrm{r}=2 \mathrm{~cm}$$ 
with gravitational potential $$\mathrm{V}=10 \mathrm{~J} / \mathrm{kg}$$. Find the gravitational potential at $$\mathrm{r}=3 \mathrm{~cm}$$ in SI unit (Given, that $$\mathrm{K}=6 \mathrm{~Jcm} / \mathrm{kg}$$)

Accepted Answer

$$E =  - {K \over {{r^2}}}$$
$$\Delta V =  - \int\limits_{r = 2\,cm}^{3\,cm} {E.\,dr} $$
$$ = \int\limits_2^3 {{k \over {{r^2}}}dr} $$
$$ = \left[ { - {K \over r}} \right]_2^3 = \left( {{K \over 6}} \right) = {6 \over 6} = 1$$ J/kg
$${V_f} - {V_i} = 1$$
$$ \Rightarrow {V_f} - 10 = 1$$
$${V_f} = 11$$ J/kg

Answer

9

Answer

11

Answer

10

Answer

12

Gravitation

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