The gravitational potential in a region is given by $V = 20 N k g^{- 1}$ (x + Y). (a) show that the equation is dimensionally correct. (b) Find the gravitational field at the point (x,y), Leave your answer in terms of the unit vectors $\to i, \to j, \to K .$ N $k g^{- 1}$ (c) Calculate the magnitude of the gravitational force on a particle of mass 500 g placed at the origin.

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Solution

$(a) V = (\frac{20 N}{k g}) (x + y) \frac{G M}{R = \frac{M L T^{- 2}}{M}}; L o r \frac{M^{- 1} l^{3} T^{- 2} M^{1}}{L} = \frac{M L^{2} T^{- 2}}{M} o r M^{0} L^{2} T^{- 2} = M^{0} L^{2} T^{- 2} ∴ L H S = R H S (b) {\to E}_{(x, y)} = - 20 (\frac{N}{K G})^i - (\frac{20 N}{k g})^j (c) \to F = \to E m = 0.5 k g [- (\frac{20 N}{k g}) - (\frac{20 N}{k g})^j] = - (10 N)^i - (10 N)^j ∴ | \to F | = \sqrt{(100) + (100)} = 10 \sqrt{2} N$

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(a)V=(20Nkg)(x+y)GMR=MLT−2M;LorM−1l3T−2M1L=ML2T−2MorM0L2T−2=M0L2T−2∴LHS=RHS(b)→E(x,y)=−20(NKG)^i−(20Nkg)^j(c)→F=→Em=0.5kg[−(20Nkg)−(20Nkg)^j]=−(10N)^i−(10N)^j∴|→F|=√(100)+(100)=10√2N