Isaac Physics

The potential at the midpoint = Potential due to moon + Potential due to planet

(edited 5 years ago)

Reply 2

OP

Original post by BobbJo

The potential at the midpoint = Potential due to moon + Potential due to planet

I got like 1.929x10^6 but its wrong

Reply 3

Original post by Kimberley_P

I got like 1.929x10^6 but its wrong

Yes it's wrong. Post the working
Potential varies inversely as distance so the formula is

\phi =- \dfrac{GM}{r}

make sure to use correct formula

Reply 4

OP

i did ((-2.4x10^22*6.674x10^-11)/2.5x10^8)+((--7.2x10^24*6.674x10^-11)/(2.5x10^8/2))
==-3.848x10^6 but still its wrong. so could you help me please???

Reply 5

Original post by Kimberley_P

i did ((-2.4x10^22*6.674x10^-11)/2.5x10^8)+((--7.2x10^24*6.674x10^-11)/(2.5x10^8/2))
==-3.848x10^6 but still its wrong. so could you help me please???

first one should be 1.25 x 10^8 in the denominator

Reply 6

OP

Original post by BobbJo

first one should be 1.25 x 10^8 in the denominator

Thank You

Reply 7

OP

HI, I have another question related to this which I can't solve

Calculate the gravitational potential at a point 6.800×10^8m from the centre of the planet in question F5.16 on the same side of the planet as its moon. You should take the Universal gravitational constant to be 𝐺=6.674×10^−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.]

F5.16 A 2.400×10^22 kg moon orbits a 7.200×10^24 planet with an orbital radius of 2.500×10^8m. Calculate the gravitational potential at the point half way between the centres of the planet and its moon. You should take the Universal gravitational constant to be 𝐺=6.674×10^−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.]
=-3.857x10^6 J/kg

Reply 8

Original post by Kimberley_P

HI, I have another question related to this which I can't solve

Calculate the gravitational potential at a point 6.800×10^8m from the centre of the planet in question F5.16 on the same side of the planet as its moon. You should take the Universal gravitational constant to be 𝐺=6.674×10^−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.]

F5.16 A 2.400×10^22 kg moon orbits a 7.200×10^24 planet with an orbital radius of 2.500×10^8m. Calculate the gravitational potential at the point half way between the centres of the planet and its moon. You should take the Universal gravitational constant to be 𝐺=6.674×10^−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.]
=-3.857x10^6 J/kg

draw a diagram to show the relative positions, then the distances to use are clearer

the gravitational potential at a point 6.800×10^8m from the centre of the planet in question F5.16 on the same side of the planet as its moon = gravitational potential due to planet at a point 6.8 x 10^8m from its centre + gravitational potential due to moon at a point (6.8 - 2.5) x10^8 m from its centre

the gravitational potential at a point 6.800×10^8m from the centre of the planet in question F5.16 on the same side of the planet as its moon = gravitational potential due to planet at a point 6.8 x 10^8m from its centre + gravitational potential due to moon at a point 4.3 x10^8 m from its centre

Reply 9

OP