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Expression for acceleration due to at height |
Answer» <html><body><p><img align="absmiddle" alt="Solution To Your Question ↓ Its value near the surface of the earth is 9.8 ms-2. Therefore, the acceleration due to gravity (g) is given by = GM/r2. 😁? Hope You Liked It 😁? Mark Me as the brainliest and Follow Me 💗?" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%3C%2Fp%3E%3Cp%3E%20%20%3C%2Fp%3E%3Cp%3E%09%09%3Cdiv%20style%3D%22%3C%2Fp%3E%3Cp%3E%09%09font-family%3A%20Cursive%3B%3C%2Fp%3E%3Cp%3E%09%09background-image%3A%20radial-gradient%28circle%20at%2077%25%2026%25%2C%20rgba%28175%2C%20175%2C%20175%2C0.06%29%200%25%2C%20rgba%28175%2C%20175%2C%20175%2C0.06%29%204%25%2Ctransparent%204%25%2C%20transparent%20100%25%29%2Cradial-gradient%28circle%20at%2037%25%2090%25%2C%20rgba%28129%2C%20129%2C%20129%2C0.06%29%200%25%2C%20rgba%28129%2C%20129%2C%20129%2C0.06%29%2055%25%2Ctransparent%2055%25%2C%20transparent%20100%25%29%2Cradial-gradient%28circle%20at%207%25%2092%25%2C%20rgba%2853%2C%2053%2C%2053%2C0.06%29%200%25%2C%20rgba%2853%2C%2053%2C%2053%2C0.06%29%2039%25%2Ctransparent%2039%25%2C%20transparent%20100%25%29%2Cradial-gradient%28circle%20at%2065%25%2017%25%2C%20rgba%28128%2C%20128%2C%20128%2C0.06%29%200%25%2C%20rgba%28128%2C%20128%2C%20128%2C0.06%29%2060%25%2Ctransparent%2060%25%2C%20transparent%20100%25%29%2Cradial-gradient%28circle%20at%20100%25%2039%25%2C%20rgba%2875%2C%2075%2C%2075%2C0.06%29%200%25%2C%20rgba%2875%2C%2075%2C%2075%2C0.06%29%2020%25%2Ctransparent%2020%25%2C%20transparent%20100%25%29%2Cradial-gradient%28circle%20at%2092%25%2034%25%2C%20rgba%28205%2C%20205%2C%20205%2C0.06%29%200%25%2C%20rgba%28205%2C%20205%2C%20205%2C0.06%29%2035%25%2Ctransparent%2035%25%2C%20transparent%20100%25%29%2Cradial-gradient%28circle%20at%2063%25%2090%25%2C%20rgba%2898%2C%2098%2C%2098%2C0.06%29%200%25%2C%20rgba%2898%2C%2098%2C%2098%2C0.06%29%2062%25%2Ctransparent%2062%25%2C%20transparent%20100%25%29%2Cradial-gradient%28circle%20at%2093%25%2074%25%2C%20rgba%28130%2C%20130%2C%20130%2C0.06%29%200%25%2C%20rgba%28130%2C%20130%2C%20130%2C0.06%29%2065%25%2Ctransparent%2065%25%2C%20transparent%20100%25%29%2Clinear-gradient%2890deg%2C%20rgb%2818%2C%20233%2C%2078%29%2Crgb%28160%2C%20239%2C%2092%29%29%3B%3C%2Fp%3E%3Cp%3E%09%09%20%20padding%3A50px%205px%3B%3C%2Fp%3E%3Cp%3E%09%09%20%20text-align%3A%20center%3B%3C%2Fp%3E%3Cp%3E%09%09%20%20font-family%3AMonoscape%3B%3C%2Fp%3E%3Cp%3E%09%09%22%3E%3C%2Fp%3E%3Cp%3E%09%09%09%3C%2Fp%3E%3Cp%3E%09%3Cspan%20style%3D%22text-align%3A%20center%3B%20border-bottom%3A2px%20double%20%23460eff%3B%09color%3A%20%23460effff%3B%20%20%22%3ESolution%20To%20Your%20Question%20%26%23x2193%3B%3C%2Fspan%3E%3Cbr%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3E%09%09%09%3C%2Fp%3E%3Cp%3E%09%3Cp%20style%3D%22color%3A%20white%3B%3C%2Fp%3E%3Cp%3E%09%20%20%20%20text-shadow%3A0px%200px%2010px%20%23eb7c7c%3B%3C%2Fp%3E%3Cp%3E%09%09%09padding%3A%2020px%3B%3C%2Fp%3E%3Cp%3E%09%09%09text-align%3A%20center%3B%3C%2Fp%3E%3Cp%3E%09%09%09font-weight%3A%20400%3B%20%22%3E%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3EIts%20value%20near%20the%20surface%20of%20the%20earth%20is%209.8%20ms-2.%20Therefore%2C%20the%20acceleration%20due%20to%20gravity%20%28g%29%20is%20given%20by%20%3D%20GM%2Fr2.%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%09%3Cbr%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3E%09%20%26%23128513%3B%20Hope%20You%20Liked%20It%09%26%23128513%3B%3C%2Fp%3E%3Cp%3E%09%3C%2Fp%3E%3C%2Fp%3E%3Cp%3E%09%3C%2Fp%3E%3Cp%3E%09%3Cmarquee%20style%3D%22color%3A%23fff%3B%22%3E%20%3Cb%3EMark%20Me%20as%20the%20brainliest%20and%20Follow%20Me%20%26%23128151%3B%3C%2Fp%3E%3Cp%3E%09%3Cbr%3E%3C%2Fp%3E%3Cp%3E%09%3C%2Fb%3E%3C%2Fmarquee%3E%3C%2Fp%3E%3Cp%3E%09%3C%2Fp%3E%3Cp%3E%09%3C%2Fp%3E%3Cp%3E%09%3C%2Fp%3E%3Cp%3E%09%09%3C%2Fdiv%3E%3C%2Fp%3E%3Cp%3E%09%3C%2Fp%3E%3Cp%3E%09%3C%2Fp%3E%3Cp%3E%09%20" title="Solution To Your Question ↓ Its value near the surface of the earth is 9.8 ms-2. Therefore, the acceleration due to gravity (g) is given by = GM/r2. 😁? Hope You Liked It 😁? Mark Me as the brainliest and Follow Me 💗?"/></p><p></p></body></html> | |