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The distance between the earth and the moon is 3.85 × 10 8 metre.At what point in between the two will the gravitational field intensity be zero? Mass of the earth = 6.0 × 10 24 k g, mass of the moon = 7.26 × 10 22 k g.
Mass of earth = M. Gravitational force constant = G. Acceleration due to gravity = g. Radius of earth = R. Now, from Newton’s law of gravitation. F = G m M R 2 E. m g = G m M R 2 E. M = g R 2 E G. Hence, the mass of the earth by using the formula is g R 2 E G
27 Απρ 2023 · question. Answer: The value in milli gram is. = 6.0 x 10³⁰ milligrams. Explanation: Using conversion factors, we can convert the mass of the Earth from teragrams to milligrams. 10³ milligrams make up one gram, which is what one teragram (10¹² grams) weighs. So we can put up the conversion factor as follows: 1 teragram = 10¹² grams ...
a sphere of mass 1kg is placed on the surface of the earth calculate the magnitude of the forceof gravitation exerted on the sphere by the earth given mass of the earth=6.0×10000000000000000000000000 kg and radius of the earth=6.4×10000000
3 Μαΐ 2020 · Mass of Earth is 5.97 × 10²⁴ kg kg and mass of the Moon is 7.35 × 10²² kg. To find : The total mass of the Earth and Moon. Solution : Step 1 of 2 : Write down mass of the Earth and Moon. Here it is given that mass of Earth is 5.97 × 10²⁴ kg and mass of Moon is 7.35 × 10²² kg. Step 2 of 2 : Find total mass of the Earth and Moon
The value of acceleration due to gravity (g) at height h above the surface of earth is given by g ′ = g R 2 (R + h) 2. If h < < R , then prove that g ′ = g ( 1 − 2 h R ) . View Solution
If the mass of moon is M / 81 where M is the mass of earth, find the distance of the point from the moon, where gravitational field due to earth and moon cancel each other. Given that distance between earth and moon is 60 R where R is the radius of earth.
Assuming earth to be a solid sphere, its moment of inertia I = 2 5 M R 2 As no external torque is acting on earth, thus its angular momentum is constant, i.e, I w = c o n s t a n t ∴ 2 5 M R 2 × 2 π T = c o n s t a n t where T is the time period of earth's rotation
Calculate acceleration due to gravity at a height of 2000 k m above the Earth's surface. Mass of Earth = 6 × 10 24 k g, radius of Earth = 6.4 × 10 6 m and (Gravitational constant, G = 6.7 × 10 − 11 N m 2 k g − 2)
5 Σεπ 2020 · Let an object of mass m is placed on the surface of earth, due to earth's gravitational field, a force of magnitude mg acting along the centre of earth. i.e., weight of body = earth gravitational force acting on the object. ⇒ mg = GMm/R². where, M is mass of earth; R is the radius of earth. i.e., R = 6370 km = 6.37 × 10⁶ m [ given ]
